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Chapter 21 Dual Nature of Radiation and Matter free study material by TEACHING CARE online tuition and coaching classes

Chapter 21 Dual Nature of Radiation and Matter free study material by TEACHING CARE online tuition and coaching classes

 

 

Electric Discharge Through Gases.

At normal atmospheric pressure, the gases are poor conductor of electricity. If we establish a potential difference (of the order of 30 kV) between two electrodes placed in air at a distance of few cm from each other, electric conduction starts in the form of sparks.

The passage of electric current through air is called electric discharge through the air.

The discharge of electricity through gases can be systematically studied with the help of discharge tube shown below

 

 

Length of discharge tube » 30 to 40 cm

Diameter of the tube » 4cm

 

 

 

 

 

The discharge tube is filled with the gas through which discharge is to be studied. The pressure of the enclosed gas can be reduced with the help of a vacuum pump and it’s value is read by manometer.

Sequence of phenomenon

As the pressure inside the discharge tube is gradually reduced, the following is the sequence of phenomenon that are observed.

 

  • At normal pressure no discharge takes
  • At the pressure 10 mm of Hg, a zig-zag thin red spark runs from one electrode to other and cracking sound is
  • At the pressure 4 of Hg, an illumination is observed at the electrodes and the rest of the tube appears dark. This type of discharge is called dark discharge.
  • When the pressure falls below 4 mm of Hg then the whole tube is filled with bright light called positive column and colour of light depends upon the nature of gas in the tube as shown in the following

 

 

 

 

Gas Colour
Air Purple red
H2 Blue
N2 Red
Cl2 Green
CO2 Bluish white
Na Yellow
Neon Dark red

 

  • At a pressure of 65 mm of Hg :
    • Sky colour light is produced at the cathode it is called as negative
    • Positive column shrinks towards the anode and the dark space between positive column and negative glow is called Faradays dark space (FDS)
  • At a pressure of 0.8 mm Hg : At this pressure, negative glow is detached from the cathode and moves towards the The dark space created between cathode and negative glow is called as Crook’s dark space length of positive column further reduced. A glow appear at cathode called cathode glow.
  • At a pressure of 05 mm of Hg : The positive column splits into dark and bright disc of light called striations.
  • At the pressure of 0.01 or 10–2 mm of Hg some invisible particle move from cathode which on striking with the glass tube of the opposite side of cathode cause the tube to These invisible rays emerging from cathode are called cathode rays.
  • Finally when pressure drops to nearly 10–4 mm of Hg, there is no discharge in

Cathode Rays.

Cathode rays, discovered by sir Willium Crooke are the stream of electrons. They can be produced by using a discharge tube containing gas at a low pressure of the order of 10–2 mm of Hg. At this pressure the gas molecules ionise and the emitted electrons travel towards positive potential of anode. The positive ions hit the cathode to cause emission of electrons from cathode. These electrons also move towards anode. Thus the cathode rays in the discharge tube are the electrons produced due to ionisation of gas and that emitted by cathode due to collision of positive ions.

(1)  Properties of cathode rays

  • Cathode rays travel in straight lines (cast shadows of objects placed in their path)
  • Cathode rays emit normally from the cathode Their direction is independent of the position of the anode.

 

 

 

  • Cathode rays exert mechanical force on the objects they
  • Cathode rays produce heat when they strikes a material
  • Cathode rays produce
  • When cathode rays strike a solid object, specially a metal of high atomic weight and high melting point X– rays are emitted from the
  • Cathode rays are deflected by an electric field and also by a magnetic
  • Cathode rays ionise the gases through which they are
  • Cathode rays can penetrate through thin foils of

 

 

  • Cathode rays are found to have velocity ranging

1 th to

30

1 th

10

of velocity of light.

 

(2)  J.J. Thomson’s method to determine specific charge of electron

It’s working is based on the fact that if a beam of electron is subjected to the crossed electric field E and

magnetic field B , it experiences a force due to each field. In case the forces on the electrons in the electron beam due to these fields are equal and opposite, the beam remains undeflected.

C = Cathode, A = Anode, F = Filament, LT = Battery to heat the filament, V = potential difference to accelerate the electrons, SS’ = ZnS coated screen, XY = metallic plates (Electric field produced between them)

  • When no field is applied, the electron beam produces illuminations at point P.
  • In the presence of any field (electric and magnetic) electron beam deflected up or down (illumination at P‘or P‘ ‘ )
  • If both the fields are applied simultaneously and adjusted such that electron beam passes undeflected and produces illumination at point P.

In this case; Electric force = Magnetic force Þ eE = evB Þ v = E ; v = velocity of electron

B

As electron beam accelerated from cathode to anode its potential energy at the cathode appears as gain in the

K.E. at the anode. If suppose V is the potential difference between cathode and anode then, potential energy = eV

 

 

And gain in kinetic energy at anode will be K.E.

= 1 mv 2

2

i.e. eV = 1 mv 2

2

Þ e m

v 2

2V

Þ e =

m

E 2

 

2VB 2

 

 

Thomson found,

e = 1.77 ´ 1011 C / kg. m

 

 

 

 

Note : @The deflection of an electron in a purely electric field is given by

y 1 æ eE ö. l 2

 

; where l length of

 

ç      ÷    2

2 è m ø v

each plate, y = deflection of electron in the field region, v = speed of the electron.

 

 

®

E

 

Positive Rays.

Positive rays are sometimes known as the canal rays. These were discovered by Goldstein. If the cathode of a discharge tube has holes in it and the pressure of the gas is around 10–3 mm of

Hg then faint luminous glow comes out from each hole on the backside of the cathode. It is said positive rays which are coming out from the holes.

 

(1)  Origin of positive rays

When potential difference is applied across the electrodes, electrons are emitted from the cathode. As they move towards anode, they gain energy. These energetic electrons when collide with the atoms of the gas in the discharge tube, they ionize the atoms. The positive ions so formed at various places between cathode and anode, travel towards the cathode. Since during their motion, the positive ions when reach the cathode, some pass through the holes in the cathode. These streams are the positive rays.

(2)  Properties of positive rays

  • These are positive ions having same mass if the experimental gas does not have However if the gas has isotopes then positive rays are group of positive ions having different masses.
  • They travels in straight lines and cast shadows of objects placed in their path. But the speed of the positive rays is much smaller than that of cathode
  • They are deflected by electric and magnetic fields but the deflections are small as compared to that for cathode
  • They show a spectrum of Different positive ions move with different velocities. Being heavy, their velocity is much less than that of cathode rays.
  • q /m ratio of these rays depends on the nature of the gas in the tube (while in case of the cathode rays q/m

is constant and doesn’t depend on the gas in the tube). q/m for hydrogen is maximum.

  • They carry energy and The kinetic energy of positive rays is more than that of cathode rays.
  • The value of charge on positive rays is an integral multiple of electronic
  • They cause ionisation (which is much more than that produced by cathode rays).

 

 

Mass Spectrograph.

It is a device used to determine the mass or (q/m) of positive ions.

(1)  Thomson mass spectrograph

 

 

 

It is used to measure atomic masses of various isotopes in gas. This is done by measuring q/m of singly ionised positive ions of the gas.

The positive ions are produced in the bulb at the left hand side. These ions are accelerated towards cathode. Some of the positive ions pass through the fine hole in the cathode. This fine ray of positive ions is subjected to

v

electric field E and magnetic field B and then allowed to strike a fluorescent screen ( E || B but E or B ^ r ).

 

If the initial motion of the ions is in + x

direction and electric and magnetic fields are applied along

  • y axis

 

then force due to electric field is in the direction of y-axis and due to magnetic field it is along z-direction.

 

The deflection due to electric field alone y = qELD

mv2

………(i)

 

The deflection due to magnetic field alone From equation (i) and (ii)

z qBLD

mv

………(ii)

 

ç      ÷

z 2 = kæ q öy , where

m

k = BLD ; This is the equation of parabola. It means all the charged particles moving

E

 

è    ø

with different velocities but of same q/m value will strike the screen placed in yz plane on a parabolic track as shown in the above figure.

Note : @All the positive ions of same. q/m moving with different velocity lie on the same parabola. Higher is the velocity lower is the value of y and z. The ions of different specific charge will lie on different

 

parabola.

Y

V   V3 V4

 

V >V >V >V

 

Light mass

q/m

q/m

q/m

q/m

 

V

2                       1        2        3       4

1

Z

Heavy mas

 

 

 

@    The number of parabola tells the number of isotopes present in the given ionic beam.

(2)  Bainbridge mass spectrograph

In Bainbridge mass spectrograph, field particles of same velocity are selected by using a velocity selector and then they are subjected to a uniform magnetic field perpendicular to the velocity of the particles. The particles corresponding to different isotopes follow different circular paths as shown in the figure.

 

 

 

  • Velocity selector : The positive ions having a certain velocity v gets isolated from all other velocity In this chamber the electric and magnetic fields are so balanced that the particle moves undeflected. For

this the necessary condition is v = E .

B

  • Analysing chamber : In this chamber magnetic field B is applied perpendicular to the direction of motion of the As a result the particles move along a circular path of

radius

 

r = mE Þ q =

qBB‘     m

E BBr

also

r1 = m1 r2            m2

 

In this way the particles of different masses gets deflected on circles of different radii and reach on different points on the photo plate.

 

 

Note : @ Separation between two traces = d = 2r2

– 2r1

Þ d = 2v(m2 – m1 ) .

qB

 

Matter waves (de-Broglie Waves).

According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle. or

A wave is associated with moving material particle which control the particle in every respect.

The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity.

 

(1)

de-Broglie wavelength

According to de-Broglie theory, the wavelength of de-Broglie wave is given by

 

l h   =       h     

Þ l µ 1 µ 1 µ   1  

 

p      mv                                                       p       v

Where h = Plank’s constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle. The smallest wavelength whose measurement is possible is that of g -rays.

The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron,

a-particle etc. is of the order of 10 -10 m.

(i)  de-Broglie wavelength associated with the charged particles.

 

 

The energy of a charged particle accelerated through potential difference V is

E = 1 mv2 = qV

2

 

 

 

Hence de-Broglie wavelength l = h =            =

p

 

 

lelectron

12.27 Å,

lproton

0.286 Å,

ldeutron

= 0.202 ´ 10 -10 Å,

laparticle

0.101 Å

 

 

(ii)  de-Broglie wavelength associated with uncharged particles.

 

 

For Neutron de-Broglie wavelength is given as l

Neutron =

m     0.286  Å

 

 

Energy of thermal neutrons at ordinary temperature

 

Q E = kT Þ l =               ; where k = Boltzman’s constant = 1.38 ´ 10 -23 Joules/kelvin , T = Absolute temp.

 

 

So l

Thermal Neutron =

30.83 Å

 

 

(2)

Some graphs

 

 

Note : @      A photon is not a material particle. It is a quanta of energy.

@    When a particle exhibits wave nature, it is associated with a wave packet, rather then a wave.

(3)  Characteristics of matter waves

  • Matter wave represents the probability of finding a particle in
  • Matter waves are not electromagnetic in
  • de-Brogile or matter wave is independent of the charge on the material It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
  • Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles is
  • Electron microscope works on the basis of de-Broglie
  • The electric charge has no effect on the matter waves or their
  • The phase velocity of the matter waves can be greater than the speed of the
  • Matter waves can propagate in vacuum, hence they are not mechanical

 

 

 

  • The number of de-Broglie waves associated with nth orbital electron is
  • Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with the orbital

(4)  Davision and Germer experiment

It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

 

 

 

 

 

 

 

 

 

The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the applied voltage to the electron gun.

According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering.

 

 

 

 

 

 

44 V

48 V

54 V

64 V

 

 

Intensity is maximum at 54 V potential difference and 50o diffraction angle.

If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula

2d sinq = nl , we can determine the wavelength of these waves.

 

Where d = distance between diffracting planes, for incident beam = Bragg’s angle.

q = (180  f) = glancing angle

2

 

q=65°

D

 

f =50°

 

The distance between diffraction planes in Ni-crystal for this experiment is d = 0.91Å and the Bragg’s angle = 65o. This gives for n = 1, l = 2 ´ 0.91 ´ 10 -10 sin 65o = 1.65 Å

q   d

Atomic planes

 

Now the de-Broglie wavelength can also be determined by using the formula l = 12.27 = 12.27 = 1.67 Å . Thus the de-Broglie hypothesis is verified.

 

 

 

 

 

Heisenberg Uncertainty Principle.

According to Heisenberg’s uncertainty principle, it is impossible to measure simultaneously both the position and the momentum of the particle.

Let Dx and Dp be the uncertainty in the simultaneous measurement of the position and momentum of the

 

particle, then

DxDp = h ; where

h = h

2p

and h = 6.63 ´ 10–34 J-s is the Planck’s constant.

 

If Dx = 0 then Dp = ¥

and if Dp = 0 then Dx = ¥ i.e., if we are able to measure the exact position of the particle (say an electron) then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly, if we are able to measure the exact linear momentum of the particle i.e., Dp = 0, then we can not measure the exact position of the particle at that time.

Photon.

According to Eienstein’s quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy.

(1)  Energy of photon

Energy of each photon is given by E = hn = hc ; where c = Speed of light, h = Plank’s constant = 6.6 ´ 10–34 Jsec,

l

n = Frequency in Hz, l = Wavelength of light

 

 

Energy of photon in electron volt

 

(2)  Mass of photon

E(eV) = hc

el

= 12375

l (Å)

» 12400

l (Å)

 

Actually rest mass of the photon is zero. But it’s effective mass is given as

 

 

E = mc 2 = hn

Þ m =  E

c 2

hn

c 2

= h . This mass is also known as kinetic mass of the photon

cl

 

(3)  Momentum of the photon

Momentum p = m ´ c = E = hn = h

c         c        l

 

(4)  Number of emitted photons

The number of photons emitted per second from a source of monochromatic radiation of wavelength l and

power P is given as (n) = P = P = Pl ; where E = energy of each photon

E      hn      hc

(5)  Intensity of light (I)

Energy crossing per unit area normally per second is called intensity or energy flux

 

 

 

 

 

i.e.

I = E = P

 

 

æ E = P = radiation powerö

 

At       A                         ç                                        ÷

 

t
è
ø

At a distance r from a point source of power P intensity is given by

I =    P

4pr 2

Þ I µ 1

r 2

 

 

 

Example: 1           The ratio of specific charge of an a -particle to that of a proton is                                                     [BCECE 2003]

(a) 2 : 1                           (b) 1 : 1                              (c) 1 : 2                        (d) 1 : 3

Solution : (c)        Specific charge = q ; Ratio = (q / m)a = qa ´ mp = 1 .

 

m                      (q / m)p         qp      ma        2

Example: 2           The speed of an electron having a wavelength of 1010m is                                                               [AIIMS 2002]

 

(a)

7.25 ´ 106 m/s                     (b)

6.26 ´ 106 m / s

(c)

5.25 ´ 106 m / s

(d)

4.24 ´ 106 m / s

 

Solution : (a)       By using lelectron =     h

mev

Þ v =      h mele

=         6.6 ´ 10-34          = 7.25 ´ 106 9.1 ´ 1031 ´ 1010

m /s.

 

Example: 3           In Thomson experiment of finding e/m for electrons, beam of electron is replaced by that of muons (particle with same charge as of electrons but mass 208 times that of electrons). No deflection condition in this case satisfied if                                                                                                                                 [Orissa (Engg.) 2002]

  • B is increased 208 times (b) E is increased 208 times

(c) B is increased 14.4 times                                             (d) None of these

e          E 2

 

Solution : (c)        In the condition of no deflection

 

= 14.4 times.

m = 2VB 2 . If m is increased to 208 times then B should be increased by

 

Example: 4           In a Thomson set-up for the determination of e/m, electrons accelerated by 2.5 kV enter the region of crossed

 

electric and magnetic fields of strengths

3.6 ´ 104 Vm1

and

1.2 ´ 103 T

respectively and go through

 

undeflected. The measured value of e/m of the electron is equal to                                                        [AMU 2002]

 

(a)

1.0 ´ 1011 C-kg-1           (b)

1.76 ´ 1011 C-kg-1           (c)

1.80 ´ 1011 C-kg-1      (d)

1.85 ´ 1011 C-kg-1

 

e          E 2          e                      (3.6 ´ 104 )2                             11

 

Solution : (c)        By using

m = 2VB 2

Þ m = 2 ´ 2.5 ´ 103 ´ (1.2 ´ 10 3 )2

= 1.8 ´ 10

C / kg.

 

 

 

Example: 5           In Bainbridge mass spectrograph a potential difference of 1000 V is applied between two plates distant 1 cm apart and magnetic field in B = 1T. The velocity of undeflected positive ions in m/s from the velocity selector is                                                                                                                                           [RPMT 1998]

 

(a) 107 m /s

E

 

(b)

V

104 m /s

1000           5

 

(c) 105 m /s

105         5

 

(d)

102 m /s

 

Solution : (c)        By using v = B ; where

E = d

= 1 ´ 10 – 2 = 10

V / m

Þ v =

1

= 10

m /s .

 

Example: 6           An electron and a photon have same wavelength. It p is the momentum of electron and E the energy of photon. The magnitude of p/ E in S.I. unit is

(a) 3.0 ´ 108                     (b) 3.33 ´ 10–9                       (c) 9.1 ´ 10–31                  (d)  6.64 ´ 10–34

 

 

Solution : (b)

l = h

p

(for electron)             or

p = h

l

and E = hc

l

(for photon)

 

\       p = 1 =             1

 

= 3.33 ´ 10 9 s / m

 

E       c      3 ´ 108 m / s

Example: 7           The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let l1 be the de-Broglie wavelength of the proton and l2 be the wavelength of the photon. The ratio l1/l2 is proportional to

[UPSEAT 2003; IIT-JEE (Screening) 2004]

 

  • E 0

E1 / 2

E -1

E -2

 

Solution : (b)       For photon l 2 = hc

E

……. (i)       and     For proton l1 =      h      

…….(ii)

 

 

Therefore

l1 =

l2

E1 / 2

Þ   l1

l2

µ E1 / 2 .

 

Example: 8           The de-Broglie wavelength of an electron having 80eV of energy is nearly ( 1eV = 1.6 ´ 1019 J , Mass of

electron 9 ´ 1031kg and Plank’s constant 6.6 ´ 1034 J-sec)                                           [EAMCET (Engg.)   2001]

(a) 140 Å                                 (b) 0.14 Å                                    (c) 14 Å                               (d) 1.4 Å

 

Solution : (d)       By using l =

= 12.27 . If energy is 80 eV then accelerating potential difference will be 80 V. So

 

 

l =              = 1.37 » 1.4 Å.

 

Example: 9           The kinetic energy of electron and proton is 1032 J . Then the relation between their de-Broglie wavelengths is                                                                                                                                           [CPMT 1999]

 

(a)

lp < le

(b)

lp > le

(c)

lp = le

(d)

lp = 2le

 

 

Solution : (a)       By using l =

 

Since mp  > me

E = 10–32 J = Constant for both particles. Hence l µ   1

so l p < le .

 

Example: 10         The energy of a proton and an a particle is the same. Then the ratio of the de-Broglie wavelengths of the proton and the a is                                                                                                                  [RPET 1991]

(a) 1 : 2                           (b) 2 : 1                              (c) 1 : 4                        (d) 4 : 1

 

Solution : (b)       By using l =      h      

Þ l µ   1  

(E – same) Þ  l proton   =

laparticle

= 2 .

1

 

Example: 11         The de-Broglie wavelength of a particle accelerated with 150 volt potential is 1010 m. If it is accelerated by 600 volts p.d., its wavelength will be                                                                                                                [RPET 1988]

 

 

 

(a) 0.25 Å                                (b) 0.5 Å                                      (c) 1.5 Å                              (d) 2 Å

 

Solution : (b)       By using l µ   1

Þ   l1 =

l2

Þ 10 -10 =

l 2

= 2 Þ l2 = 0.5 Å.

 

Example: 12         The de-Broglie wavelength of an electron in an orbit of circumference 2pr is                                       [MP PET 1987]

 

(a)

2pr

(b) pr

(c)

1 / 2pr

(d)

1 / 4pr

 

Solution : (a)       According to Bohr’s theory mv r = n h            Þ 2p r = n æ h ö = nl

2p                     ç mv ÷

è        ø

For n = 1     l = 2pr

Example: 13         The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is (taking h = 6 ´ 10 34 J-sec)                                                                                            [Kerala (Engg.)  2002]

 

(a) 100                            (b) 1000                             (c)

3 ´ 10 20

(d)

3 ´ 1018

 

Solution : (c)        By using n = Pl

hc

=      100 ´ 540 ´ 10 9

6.6 ´ 10 34 ´ 3 ´ 108

= 3 ´ 1020

 

Example: 14         A steel ball of mass 1kg is moving with a velocity 1 m/s. Then its de-Broglie waves length is equal to

(a) h                                          (b) h / 2                              (c) Zero                        (d) 1 / h

 

Solution : (a)       By using l = h  

mv

Þ l =   l   = h.

1 ´ 1

 

Example: 15         The de-Broglie wavelength associated with a hydrogen atom moving with a thermal velocity of 3 km/s will be (a) 1 Å                                                    (b) 0.66 Å                                    (c) 6.6 Å                              (d) 66 Å

 

Solution : (b)       By using l =       h

mvrms

Þ l =               6.6 ´ 10 34

2 ´ 1.67 ´ 10 27 ´ 3 ´ 103

= 0.66 Å

 

Example: 16         When the momentum of a proton is changed by an amount P0, the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was                                   [CPMT 2002]

(a) p0                               (b) 100 p0                           (c) 400 p0                     (d) 4 p0

 

 

Solution : (c)

l µ 1 Þ Dp = – Dl   Þ           =

 

Þ p0 = 0.25 =    1

 

 

 

Þ p = 400 p .

 

p            p            l

p        100

400                      0

 

Example: 17         If the electron has same momentum as that of a photon of wavelength 5200Å, then the velocity of electron in

m /sec is given by

(a) 103                             (b) 1.4 ´ 103                        (c) 7 ´ 10–5                    (d) 7.2 ´ 106

 

 

Solution : (b)

l = h mv

Þ v = h ml

=                  6.6 ´ 10 34

9.1 ´ 10 31 ´ 5200 ´ 10 10

Þ v = 1.4 ´ 103

m/s.

 

Example: 18         The de-Broglie wavelength of a neutron at 27oC is l. What will be its wavelength at 927oC (a) l / 2                                       (b) l / 3                             (c) l / 4                       (d) l / 9

 

Solution : (a)        l

µ   1  

Þ l1 =

 

Þ   l =

=                 = 2

Þ l   = l .

 

neutron                                     l2                          l2                                                                        2       2

Example: 19         The de-Broglie wavelength of a vehicle is l. Its load is changed such that its velocity and energy both are doubled. Its new wavelength will be

 

 

 

 

  • l (b)

l                                            (c)

2

l                                    (d) 2l

4

 

 

Solution : (a)

l = h mv

and E = 1 mv 2

2

Þ l = hv

2E

when v and E both are doubled, l remains unchanged i.e. l‘ = l.

 

Example: 20  In Thomson mass spectrograph when only electric field of strength 20 kV/m is applied, then the displacement of the beam on the screen is 2 cm. If length of plates = 5 cm, distance from centre of plate to the screen = 20 cm and velocity of ions = 106 m/s, then q/m of the ions is

(a) 106 C/kg                             (b) 107 C/Kg                                (c) 108 C/kg                        (d) 1011 C/kg

Solution : (c)        By using y = qELD ; where y = deflection on screen due to electric field only

mv 2

 

Þ q =

m

yv 2 =

ELD

2 ´ 10 2 ´ (106 )2

 

20 ´ 103 ´ 5 ´ 10 2 ´ 0.2

= 108 C / kg.

 

Example: 21         The minimum intensity of light to be detected by human eye is  1010 W / m2 . The number of photons of

wavelength 5.6 ´ 107 m entering the eye, with pupil area 106 m2 , per second for vision will be nearly (a) 100                                       (b) 200                               (c)  300                         (d) 400

 

Solution : (c)        By using

I = P ;

A

where P = radiation power

 

 

Þ P = I ´ A

Þ nh c = IA

Þ n IAl

 

tl                  t       hc

Tricky example: 1
è
ø

Hence number of photons entering per sec the eye æ n ö = 10 -10 ´ 10 -6 ´ 5.6 ´ 10 -7 = 300.

 

ç t ÷

6.6 ´ 10 34 ´ 3 ´ 108

 

 

 

Velocity of       13

photon (c)

 

 

 

 

 

Photo-electric Effect.

It is the phenomenon of emission of electrons from the surface of metals, when light radiations (Electromagnetic radiations) of suitable frequency fall on them. The emitted electrons are called photoelectrons and the current so produced is called photoelectric current.

This effect is based on the principle of conservation of energy.

(1)  Terms related to photoelectric effect

  • Work function (or threshold energy) (W0) : The minimum energy of incident radiation, required to eject the electrons from metallic surface is defined as work function of that

 

W   = hn    = hc Joules ;

n0 = Threshold frequency; l0 = Threshold wavelength

 

l

0            0

0

Work function in electron volt W0(eV) = hc = 12375

el0        l0 (Å)

Note : @              By coating the metal surface with a layer of barium oxide or strontium oxide it’s work function is lowered.

  • Threshold frequency (n0) : The minimum frequency of incident radiations required to eject the electron from metal surface is defined as threshold

If incident frequency n < n0 Þ No photoelectron emission

  • Threshold wavelength (l0) : The maximum wavelength of incident radiations required to eject the electrons from a metallic surface is defined as threshold

If incident wavelength l > l0 Þ No photoelectron emission

(2)  Einstein’s photoelectric equation

According to Einstein, photoelectric effect is the result of one to one inelastic collision between photon and electron in which photon is completely absorbed. So if an electron in a metal absorbs a photon of energy E (= hn), it uses the energy in three following ways.

  • Some energy (say W) is used in shifting the electron from interior to the surface of the
  • Some energy (say W0) is used in making the surface electron free from the
  • Rest energy will appear as kinetic energy (K) of the emitted

 

 

 

Hence E = W + W0 + K

For the electrons emitting from surface W = 0 so kinetic energy of emitted electron will be max.

Hence          E = W0 + Kmax ;            This is the Einstein’s photoelectric equation

(3)  Experimental arrangement to observe photoelectric effect

When light radiations of suitable frequency (or suitable wavelength and suitable energy) falls on plate P, photoelectrons are emitted from P.

  • If plate Q is at zero potential r.t. P, very small current flows in the circuit because of some electrons of high kinetic energy are reaching to plate Q, but this current has no practical utility.
  • If plate Q is kept at positive potential r.t. P current starts flowing through the circuit because more electrons are able to reach upto plate Q.
  • As the positive potential of plate Q increases, current through the circuit increases but after some time constant current flows through the circuit even positive potential of plate Q is still increasing, because at this condition all the electrons emitted from plate P are already reached up to plate Q. This constant current is called saturation
  • To increase the photoelectric current further we will have to increase the intensity of incident

 

  • To decrease the photoelectric current plate Q is maintained at negative potential r.t. P, as the anode Q is made more and more negative, fewer and fewer electrons will reach the cathode and the photoelectric current decreases.
  • At a particular negative potential of plate Q no electron will reach the plate Q and the current will become zero, this negative potential is called stopping potential denoted by V0.
  • If we increase further the energy of incident light, kinetic energy of photoelectrons increases and more negative potential should be applied to stop the electrons to reach upto plate Q. Hence eV0 = Kmax.

Note : @ Stopping potential depends only upon frequency or wavelength or energy of incident radiation. It doesn’t depend upon intensity of light.

We must remember that intensity of incident light radiation is inversely proportional to the square of

 

distance between source of light and photosensitive plate P i.e.,

 

Important formulae

I µ 1

d 2

so I µ i µ 1 )

d 2

 

Þ hn

= hn 0 + Kmax

 

 

 

Þ K        = eV

= h(nn )  Þ 1 mv 2

= h(nn )  Þ v        =

 

max           0

0              2     max

0                max

 

 

 

1     2                 æ 1    1 ö        æ l0l ö

Þ Kmax = 2 mvmax = eV0 = hc ç    –      ÷ = hc ç           ÷ Þ vmax =

è l     l0 ø        è ll0   ø

h                      hc æ 1    1 ö              æ 1    1 ö

Þ V0 = e (nn 0 ) = e ç       –      ÷ = 12375 ç    –      ÷

 

 

è l

(4)  Different graphs

l0 ø

è l     l0 ø

 

  • Graph between potential difference between the plates P and Q and photoelectric current
  • Graph between maximum kinetic energy / stopping potential of photoelectrons and frequency of incident light

 

 

 

 

 

 

 

 

 

Photoelectric Cell.

A device which converts light energy into electrical energy is called photoelectric cell. It is also known as photocell or electric eye.

Photoelectric cell are mainly of three types

 

 

 

 

It consists of an evacuated glass or quartz bulb containing anode A and cathode C. The cathode is semi- cylindrical metal on which a layer of photo-sensitive material is coated.

It is based on the principle that conductivity of a semiconductor increases with increase in the intensity of incident light.

It consists of a Cu plate coated with a thin layer of cuprous oxide (Cu2O). On this plate is laid a semi transparent thin film of silver.

Transparent

 

Light    A C

 

 

Galvanometer or

Micro ammeter

Surface film

 

 

Selenium

Selenium Metal layer

 

 

R        Output

film of silver

 

A

C                                                                      R

 

 

m A

+          –

 

 

 

 

Note : @  The photoelectric current can be increased by filling some inert gas like Argon into the bulb. The photoelectrons emitted by cathode ionise the gas by collision and hence the current is increased.

Compton effect

The scattering of a photon by an electron is called Compton effect. The energy and momentum is conserved. Scattered photon will have less energy (more wavelength) as compare to incident photon (less wavelength). The energy lost by the photon is taken by electron as kinetic energy.

The change in wavelength due to Compton effect is called Compton shift. Compton shift

 

lfli

= h (1 – cosq )

m0c

 

 

 

 

 

 

 

 

Note : @ Compton effect shows that photon have momentum.

X-rays.

X-rays was discovered by scientist Rontgen that’s why they are also called Rontgen rays.

Rontgen discovered that when pressure inside a discharge tube kept 10–3 mm of Hg and potential difference is 25 kV then some unknown radiations (X-rays) are emitted by anode.

(1)  Production of X-rays

There are three essential requirements for the production of X-rays

  • A source of electron
  • An arrangement to accelerate the electrons
  • A target of suitable material of high atomic weight and high melting point on which these high speed electrons

(2)  Coolidge X-ray tube

 

 

 

It consists of a highly evacuated glass tube containing cathode and target. The cathode consist of a tungsten filament. The filament is coated with oxides of barium or strontium to have an emission of electrons even at low temperature. The filament is surrounded by a molybdenum cylinder kept at negative potential w.r.t. the target.

The target (it’s material of high atomic weight, high melting point and high thermal conductivity) made of tungsten or molybdenum is embedded in a copper block.

The face of the target is set at 45o to the incident electron stream.

The filament is heated by passing the current through it. A high potential difference (» 10 kV to 80 kV) is applied between the target and cathode to accelerate the electrons which are emitted by filament. The stream of highly energetic electrons are focussed on the target.

Most of the energy of the electrons is converted into heat (above 98%) and only a fraction of the energy of the electrons (about 2%) is used to produce X-rays.

During the operation of the tube, a huge quantity of heat is produced in this target, this heat is conducted through the copper anode to the cooling fins from where it is dissipated by radiation and convection.

  • Control of intensity of X-rays : Intensity implies the number of X-ray photons produced from the target. The intensity of X-rays emitted is directly proportional to the electrons emitted per second from the filament and this can be increased by increasing the filament So intensity of X-rays µ Filament current
  • Control of quality or penetration power of X-rays : Quality of X-rays implies the penetrating power of X-rays, which can be controlled by varying the potential difference between the cathode and the

For large potential difference, energy of bombarding electrons will be large and hence larger is the penetration power of X-rays.

Depending upon the penetration power, X-rays are of two types

 

Note : @ Production of X-ray is the reverse phenomenon of photoelectric effect.

(3)  Properties of X-rays

  • X-rays are electromagnetic waves with wavelength range 1Å – 100Å.
  • The wavelength of X-rays is very small in comparison to the wavelength of Hence they carry much more energy (This is the only difference between X-rays and light)

 

 

 

  • X-rays are
  • They travel in a straight line with speed of
  • X-rays are measured in Rontgen (measure of ionization power).
  • X-rays carry no charge so they are not deflected in magnetic field and electric
  • lGama rays <lXrays lUV rays
  • They used in the study of crystal
  • They ionise the gases
  • X-rays do not pass through heavy metals and
  • They affect photographic
  • Long exposure to X-rays is injurious for human
  • Lead is the best absorber of X-rays.
  • For X-ray photography of human body parts, BaSO4 is the best
  • They produce photoelectric effect and Compton effect
  • X-rays are not emitted by hydrogen
  • These cannot be used in Radar because they are not reflected by the
  • They show all the important properties of light rays like; reflection, refraction, interference, diffraction and polarization etc.

(4)  Absorption of X-rays

X-rays are absorbed when they incident on substance.

 

Intensity of emergent X-rays

I = I 0 e mx

 

So intensity of absorbed X-rays

I‘ = I 0

  • I = I 0

(1 – e mx )

 

where x = thickness of absorbing medium, m = absorption coefficient

Note : @ The thickness of medium at which intensity of emergent X-rays becomes half i.e.

I‘ = I 0

2

 

is called

 

half value thickness (x1/2) and it is given as

 

 Classification of X-rays.

x1 / 2

0.693 .

m

 

 

In X-ray tube, when high speed electrons strikes the target, they penetrate the target. They loses their kinetic energy and comes to rest inside the metal. The electron before finally being stopped makes several collisions with the atoms in the target. At each collision one of the following two types of X-rays may get form.

(1)  Continuous X-rays

As an electron passes close to the positive nucleus of atom, the electron is deflected from it’s path as shown in figure. This results in deceleration of the electron. The loss in energy of the electron during deceleration is emitted in the form of X-rays.

The X-ray photons emitted so form the continuous X-ray spectrum.

 

 

 

 

 

Note : @              Continuos X-rays are produced due to the phenomenon called “Bremsstrahlung”. It means slowing down or braking radiation.

Minimum wavelength

When the electron looses whole of it’s energy in a single collision with the atom, an X-ray photon of maximum

 

energy hnmax is emitted i.e.

1 mv2 = eV = hn

2

max

  hc  

l

 

min

where v = velocity of electron before collision with target atom, V = potential difference through which electron is accelerated, c = speed of light = 3 ´ 108 m/s

 

Maximum frequency of radiations (X-rays)

n max

eV

h

 

Minimum wave length = cut off wavelength of X-ray

lmin

hc eV

= 12375 Å V

 

Note : @              Wavelength of continuous X-ray photon ranges from certain minimum (lmin) to infinity.

 

 

 

nmax

logenmax

lmin

logelmax

 

 

 

 

V                                                    logeV

Intensity wavelength graph

V                                                        logeV

 

Y

 

The continuous X-ray spectra consist of all the wavelengths over a given range. These wavelength are of different intensities. Following figure shows the intensity variation of different wavelengths for various accelerating voltages applied to X-ray tube.

 

 

 

lmin

 

 

 

 

Wave length

 

30 kV

20 kV

10 kV

 

For each voltage, the intensity curve starts at a particular minimum wavelength (lmin). Rises rapidly to a maximum and then drops gradually.

The wavelength at which the intensity is maximum depends on the accelerating voltage, being shorter for higher voltage and vice-versa.

(2)  Characteristic X-rays

Few of the fast moving electrons having high velocity penetrate the surface atoms of the target material and knock out the tightly bound electrons even from the inner most shells of the atom. Now when the electron is knocked out, a vacancy is created at that place. To fill this vacancy electrons from higher shells jump to fill the

 

 

 

created vacancies, we know that when an electron jumps from a higher energy orbit E1 to lower energy orbit E2, it radiates energy (E1E2). Thus this energy difference is radiated in the form of X-rays of very small but definite wavelength which depends upon the target material. The X-ray spectrum consist of sharp lines and is called characteristic X-ray spectrum.

 

 

 

 

 

 

 

 

 

K, L, M, …… series

If the electron striking the target eject an electron from the K-shell of O the atom, a vacancy is crated in the K-shell. Immediately an electron from N one of the outer shell, say L-shell jumps to the K-shell, emitting an X-ray M

photon of energy equal to the energy difference between the two shells. Similarly, if an electron from the M-shell jumps to the K-shell, X-ray   L

 

 

n=5 n=4

 

     
      Ma Mb  
    La Lb Lg M-series
Ka Kb Kg L-series

 

n=3 n=2

 

photon of higher energy is emitted. The X-ray photons emitted due to the

jump of electron from the L, M, N shells to the K-shells gives K , K , K

K                                                                                                 n=1

K-series

 

lines of the K-series of the spectrum.

a          b          g

 

If the electron striking the target ejects an electron from the L-shell of the target atom, an electron from the M, N ….. shells jumps to the L-shell so that X-rays photons of lesser energy are emitted. These photons form the lesser energy emission. These photons form the L-series of the spectrum. In a similar way the formation of M series, N series etc. may be explained.

Energy and wavelength of different lines

 

Note : @              The wavelength of characteristic X-ray doesn’t depend on accelerating voltage. It depends on the atomic number (Z) of the target material.

 

@           lKa

< lLa

< lMa

and n Ka

  • nLa
  • nMa

Intensity

 

@           lKa

>  lLb

< lKg

 

Intensity-wavelength graph

At certain sharply defined wavelengths, the intensity of X-rays is very

 

 

lmin                                   Wavelength

 

 

 

large as marked Ka, Kb …. As shown in figure. These X-rays are known as characteristic X-rays. At other wavelengths the intensity varies gradually and these X-rays are called continuous X-rays.

 

 

Mosley’s law

Mosley studied the characteristic X-ray spectrum of a number of a heavy elements and concluded that the spectra of different elements are very similar and with increasing atomic number, the spectral lines merely shift towards higher frequencies.

Z

 

He also gave the following relation         = a (Z b)

 

where n = Frequency of emitted line, Z = Atomic number of target, a = Proportionality constant,

b = Screening constant.

Note : @              a and b doesn’t depend on the nature of target. Different values of b are as follows

 

b = 1 for K-series
b = 7.4 for L-series
b = 19.2 for M-series

@          (Z b) is called effective atomic number.

 

 

More about Mosley’s law

  • It supported Bohr’s theory
  • It experimentally determined the atomic number (Z) of
  • This law established the importance of ordering of elements in periodic table by atomic number and not by atomic
  • Gaps in Moseley’s data for A = 43, 61, 72, 75 suggested existence of new elements which were later
  • The atomic numbers of Cu, Ag and Pt were established to be 29, 47 and 78
  • When a vacancy occurs in the K-shell, there is still one electron remaining in the K-shell. An electron in the L-shell will feel an effective charge of (Z – 1)e due to + Ze from the nucleus and – e from the remaining K-shell electron, because L-shell orbit is well outside the K-shell
    

1              2 æ 1     1 ö

 

  • Wave length of characteristic spectrum

l = R(Z b)

ç

ç n2

÷

n2 ÷

and energy of X-ray radiations.

 

    

hc                             2 æ 1     1 ö

 

è 1         2 ø

 

DE = hn = l

= Rhc(Z b)

ç

ç n2

÷

n2 ÷

 

è 1          2 ø

  • If transition takes place from n2 = 2 to n1 = 1 (Ka – line)

 

 

 

 

(a) a =            = 2.47 ´ 1015 Hz

 

 

(b)   n

= RC(Z – 1)2 æ1 –

1 ö = 3RC (Z – 1)2 = 2.47 ´ 1015 (Z – 1)2 Hz

 

Ka                                             ç

è

22 ÷       4

 

ø
  • In general the wavelength of all the K-lines are given by

1 = R(Z – 1)2 æ1 –

 

1 ö where n = 2, 3, 4, ….

 

 

While for Ka line l Ka

ø

= 1216 Å

(Z – 1)

ç

l K                           è

n2 ÷

 

EKa

= 10.2(Z – 1)2 eV

 

Uses of X-rays

  • In study of crystal structure : Structure of DNA was also determined using X-ray
  • In medical
  • In radiograph
  • In radio therapy
  • In engineering
  • In laboratories
  • In detective department
  • In art the change occurring in old oil paintings can be examined by X-rays.

 

 

 

 

 

Example: 22         The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately                                                                               [AIEEE 2004]

(a) 540 nm                              (b) 400 nm                                  (c) 310 nm                          (d) 220 nm

 

Solution : (c)        By using l0 = 12375

W0 (eV)

Þ  l0 = 12375 = 3093.7 Å  ~– 310 nm

4

 

Example: 23         Photo-energy 6 eV are incident on a surface of work function 2.1 eV. What are the stopping potential

[MP PMT 2004]

(a) – 5V                                    (b) – 1.9 V                                   (c) – 3.9 V                           (d) – 8.1 V

 

Solution : (c)        By using Einstein’s equation E = W0 + Kmax Þ

6 = 2.1 + Kmax

Þ K max = 3.9 eV

 

 

Also V0

= – Kmax

r

= – 3.9 V.

 

Example: 24 When radiation of wavelength l is incident on a metallic surface the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is                                                                            [EAMCET (Engg.) 2003]

 

  • 2l
  • 4l
  • 6l
  • 8l

 

 

Solution : (b)       By using V

hc é 1 1 ù

 

l
û
e
ë

0            ê         l0 ú

 

ê

4.8 = hc é 1 – 1 ù

 

 

…… (i)         and

1.6 = hc é 1

 

– 1 ù

 

…… (ii)

 

l
û

e ë         l0 ú

e ê 2l

l0 ú

 

ë
û

From equation (i) and (ii)

l0 = 4l.

 

Example: 25         When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m for the electron is 1.8 ´ 1011 Ckg 1 the maximum velocity of the ejected electrons is                                            [Kerala (Engg.) 2002]

 

(a)

6 ´ 105ms1

(b)

8 ´ 105ms1

(c)

1.8 ´ 106ms1

(d)

1.8 ´ 105ms1

 

 

Solution : (c)

1 mv 2     = eV       Þ   v         =                          =

 

= 1.8 ´ 106 m / s .

 

2      max           0

max

 

 

Example: 26         The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 eV) will be                                                                                            [JIPMER 2002]

 

(a)

4 ´ 1010 Hz

(b)

4 ´ 1011 Hz

(c)

4 ´ 1014 Hz

(d)

4 ´ 1010 Hz

 

Solution : (c)        Threshold wavelength l0 = 12375

W0 (eV)

= 12375 = 7500 Å.

1.65

 

 

\ so minimum frequency

n    = c   =

0       l0

3 ´ 108

7500 ´ 10 10

= 4 ´ 1014 Hz.

 

Example: 27 Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will be                                                                                                                                          [AIEEE 2002]

(a) 1 : 5                           (b) 1 : 4                              (c) 1 : 2                        (d) 1 : 1

 

 

 

 

 

Solution : (b)       By using K

= E W

Þ (K max )1 =

 

1 – 0.5

= 0.5 = 1 .

 

 

 

max

0          (K max )2

2.5 – 0.5       2      4

 

Example: 28     Photoelectric emission is observed from a metallic surface for frequencies n 1 and n 2 of the incident light rays (n1 > n 2) . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is                                                                                 [EAMCET (Engg.) 2001]

 

(a)

n1n 2

k – 1

(b)

kn1n 2

 

k – 1

(c)

kn 2n1

 

k – 1

(d)

n 2n1

 

k – 1

 

Solution : (b)       By using hnhn 0 = kmax

Þ h(n 1n 0 ) = k1 and

h(n 1n 0 ) = k 2

 

Hence n 1 – n 0

n 2n 0

= k1 = 1

k 2       k

Þ n 0

kn 1n 2

k – 1

 

Example: 29         Light of frequency

8 ´ 1015 Hz

is incident on a substance of photoelectric work function 6.125 eV. The

 

maximum kinetic energy of the emitted photoelectrons is                                                                  [AFMC 2001]

(a) 17 eV                                 (b) 22 eV                                     (c) 27 eV                             (d) 37 eV

Solution : (c)        Energy of incident photon E = hn = 6.6 ´ 10 34 ´ 8 ´ 1015 = 5.28 ´ 10 18 J = 33 eV.

 

From E = W0 + K max

Þ K max

= E W0 = 33 – 6.125 = 26.87 eV » 27 eV.

 

Example: 30         A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a distance of 2 m, then the ejected electron                                                                             [MNR 1986; UPSEAT 2000, 2001]

  • Moves with one-fourth energy as that of the initial energy
  • Moves with one fourth of momentum as that of the initial momentum
  • Will be half in number
  • Will be one-fourth in number

 

 

Solution : (d)       Number of photons µ Intensity µ

1

 

(distance)2

 

N        æ d ö 2

N        æ 2 ö 2                  N

 

Þ     1 = ç   2 ÷

Þ      1 – ç   ÷        Þ N 2 =     1 .

 

N 2      è d1 ø

N 2     è 1 ø                           4

 

Example: 31         When yellow light incident on a surface no electrons are emitted while green light can emit. If red light is incident on the surface then                                                                  [MNR 1998; MH CET 2000; MP PET 2000]

(a) No electrons are emitted                                            (b) Photons are emitted

(c) Electrons of higher energy are emitted                           (d) Electrons of lower energy are emitted

 

Solution : (a)

lGreen < lYellow < lRed

According to the question

lGreen

is the maximum wavelength for which photoelectric emission takes place.

 

Hence no emission takes place with red light.

Example: 32   When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is (h = Planck’s constant, c = velocity of light in air)                                                                                                                [EMCET (Engg.) 2000]

 

(a)

2hc ´ 106 J

(b)

1.5hc ´ 106 J

(c)

hc ´ 106 J

(d)

0.5hc ´ 106 J

 

 

Solution : (a)       By using

E = W   + K

Þ hc = W   + 1 mv 2

 

 

 

0         max

l        0     2

 

 

 

 

         hc         = W0 + 1 mv 2           ……(i)       and

         hc         = W0 + 1 m(2v)2

……(ii)

 

400 ´ 10 9               2

From equation (i) and (ii) W0 = 2hc ´ 106 J.

250 ´ 10 9               2

 

Example: 33 The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the surfaces of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B) [EAMCET (Med.) 2000]

(a) 1 : 1                           (b) 1 : 2                              (c) 1 : 3                        (d) 1 : 4

 

Solution : (b)       By using

E = W0 + K max

Þ  EA  = hf = WA  + KA

and EB = h(2 f ) = WB + KB

 

 

So,

1WA KA

……(i)      also it is given that WA  = 1

……..(ii)

 

2     WB  + KB                                                                WB            2

 

From equation (i) and (ii) we get K A

KB

= 1 .

2

 

Example: 34 When a point source of monochromatic light is at a distance of 0.2m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the photoelectric cell, then                                                                                    [IIT-JEE 1992; MP PMT 1999]

  • The stopping potential will be 2 V (b) The stopping potential will be 0.6 V

(c) The saturation current will be 6 mA                                           (d) The saturation current will be 18 mA

 

 

Solution : (b)       Photoelectric current (i) µ Intensity µ

1        . If distance becomes 0.6 m (i.e. three times) so current

(distance)2

 

 

becomes

1 times i.e. 2mA.

9

 

Also stopping potential is independent of intensity i.e. it remains 0.6 V.

Example: 35         In a photoemissive cell with exciting wavelength l , the fastest electron has speed v. If the exciting wavelength is changed to 3l / 4 , the speed of the fastest emitted electron will be                                                                      [CBSE 1998]

 

(a)

v (3 / 4)1 / 2

(b)

v (4 / 3)1 / 2

  • Lessthen v (4 / 3)1 / 2
  • Greaterthen v (4 / 3)1 / 2

 

 

 

 

Solution : (d)       From E = W

+ 1 mv 2       Þ   v         =

(where E = hc )

 

 

 

0      2     max               max                                                            l

 

 

If wavelength of incident light charges from l to

3l (decreases)

4

 

Let energy of incident light charges from E to E‘ and speed of fastest electron changes from v to v ¢ then

 

 

v =                                  …..(i) and

v‘ =

…….(ii)

 

 

 

1                  4                                                            æ 4 ö1/ 2

 

As E µ

Þ     E‘ =

E hence

v‘ =

Þ v‘ = ç   ÷

 

l                 3                                                            è 3 ø

 

 

 

 

æ 4 ö1 / 2

Þ v‘ = ç 3 ÷                  X =

æ 4 ö1 / 2

v so v‘ > ç 3 ÷         v .

 

è   ø                                                                                      è ø

 

 

 

Example: 36         The minimum wavelength of X-rays produced in a coolidge tube operated at potential difference of 40 kV is

 

(a) 0.31Å                         (b) 3.1Å                              (c) 31Å                        (d) 311Å

 

 

[BCECE 2003]

 

 

Solution : (a)

lmin =   12375 = 0.309 Å  » 0.31 Å

40 ´ 103

 

Example: 37         The X-ray wavelength of La   line of platinum (Z = 78) is 1.30 Å. The X –ray wavelength of La   line of Molybdenum

(Z = 42) is                                                                                                              [EAMCET (Engg.) 2000]

(a) 5.41Å                         (b) 4.20Å                            (c) 2.70Å                      (d) 1.35 Å

 

 

 

Solution : (a)       The wave length of

L   line is given by

1 = R(z – 7.4)2æ 1

 

– 1 ö Þ l µ         1

 

 

 

a                                 l                   ç 22

32 ÷

(z – 7.4)2

 

è
ø

Þ l1 = (z2 – 7.4)2 Þ 1.30 = (42 – 7.4)2

 

 

l = 5.41Å .

 

l2     (z1

– 7.4)2

l2      (78 – 7.4)2          2

 

 

Example: 38         The cut off wavelength of continuous X-ray from two coolidge tubes operating at 30 kV but using different target materials (molybdenum Z= 42 and tungsten Z = 74) are

(a) 1Å, 3Å                        (b) 0.3 Å, 0.2 Å                    (c) 0.414 Å, 0.8 Å          (d) 0.414 Å, 0.414 Å

Solution : (d)       Cut off wavelength of continuous X-rays depends solely on the voltage applied and does not depend on the material of the target. Hence the two tubes will have the same cut off wavelength.

 

Ve = hn = hc

l

or l = hc Ve

= 6.627 ´ 10 -34 ´ 3 ´ 108 m = 414 ´ 10 =10 m = 0.414 Å.

30 ´ 103 ´ 1.6 ´ 10 19

 

Tricky example: 1

 

 

Tricky example: 3

 

Two photons, each of energy 2.5eV are simultaneously incident on the metal surface. If the work function of the metal is 4.5 eV, then from the surface of metal

(a) Two electrons will be emitted                                      (b) Not even a single electron will be emitted

(c) One electron will be emitted                                        (d) More than two electrons will be emitted

Solution : (b) Photoelectric effect is the phenomenon of one to one elastic collision between incident photon and an electron. Here in this question one electron absorbs one photon and gets energy 2.5 eV which is lesser than 4.5 eV. Hence no photoelectron emission takes place.

Tricky example: 4
In X-ray tube when the accelerating voltage V is halved, the difference between the wavelength of Ka

line and minimum wavelength of continuous X-ray spectrum

(a) Remains constant                                                     (b) Becomes more than two times

(c) Becomes half                                                           (d) Becomes less than two times

Solution : (c)        Dl = lKa    – lmin  when V is halved lmin becomes two times but l Ka      remains the same.

\        Dl‘ = lKa    – 2lmin = 2(Dl) – lKa

\ Dl ‘ < 2(Dl)

Tricky example: 5
Molybdenum emits Ka-photons of energy 18.5 keV and iron emits Ka photons of energy 34.7 keV. The times taken by a molybdenum Ka photon and an iron Ka photon to travel 300 m are

(a) (3 ms, 15 ms)                   (b) (15 ms, 3ms)                    (c) (1 ms, 1 ms)                 (d) (1 ms, 5ms)

Solution : (c)      Photon have the same speed whatever be their energy, frequency, wavelength, and origin.

\ time of travel of either photon =           300    = 10 6 s = 1m s

3 ´ 108

 

 

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