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Atomic Structure 1 Hints and Solutions

Atomic Structure

 

 

    Hints to Subjective Problems

 

Level – I

  1. Longest wavelength is associated with least energy.
  2. Total energy emitted = number of photons emitted × energy of 1 photon
  3. At the closest distance of approach, the alpha particle will have the same kinetic and potential energies
  4. Period of revolution = T =

Level – II

  1. a) KE for a H – like atom is same as the magnitude of total energy
  2. b) l µ 1/m if the velocity is same.
  3. Use quantization condition for a Bohr orbit in order to evaluate the n values.
  4. The energy of the incident light should always be greater than the work function for photo electric emission to take place.

 

Level – III

  1. Identify the wavelengths corresponding to the transitions; which one is shorter or larger. Using the appropriate evaluate the work function and hence stopping potential.
  2. Evaluate the uncertainty in position with the help of radius of first orbit of He+ ion
  3. d) If the energy of the incoming electron is less than DE then it cannot interact with the atom and if the energy is more then transition is made from one level to another and the remaining energy will be in the form of K.E.
  4. One Rydberg = IP of H atom.
  5. Energy of incident photon will be partially utilized in breaking the I2 molecule and residual energy in the form of K.E.
  6. Removal of electron involves absorption of energy and addition of electron involves release of energy.

Energy absorbed by F + Energy absorbed by Cl = Total energy absorbed

Energy released by F + Energy released by Cl = Total energy released

  1. Using quantization condition and the condition for the stability of the electron in the circular orbit expression for r can be derived.

 

 

 

 

    Solutions to Subjective Problems

Level – I

For He; Z = 2; For Paschen series n1 = 3

For longest wavelength n2 = 4

 

= 109678×4×

= 109678 ×4×

l = 4689 Ao

 

  1. wave number of first line of Balmer,  =

\wave length of first line of Balmer =

wave  number of ultimate line of Balmer,  =  =

\wave length of  ultimate line of Balmer =

            \Ratio =

 

  1. Energy corresponding to 60 W bulb is

E = 60 J/sec = 60 ×3600 J/hr

For 10 hours the energy is

E = 60×3600×10

Energy corresponding to the photon =

= 3.37×10-19 J

\ No. of photons emitted =

                  = 6.404 ×1024

 

  1. From equation En = , the energy change of a H–atom that goes from initial state of quantum number ni to a final state of quantum number nf is

DE = Ef – Ei =

Hence ni = 1, nf = 3, E1 = –13.6 eV

DE = –13.6  = 12.08 eV.

  1. (a) From equation rn = n2ao

n =  » 435

(b) En =

(Such an atom would obviously be extremely fragile and can be easily ionized).

 

  1. At the distance of closest approach of an alpha particle (r)

P.E. = K.E.

 

v2 =

=

v= 6.35 ×106 m/sec

 

=  =

1.54 =

V = 63.24 volts

 

  1. 8. Dx × Dp ³

(Dp)2 = (Dx)2 =  (since Dp = Dx)

Dp =  =

mDv = Dp =

Dv =  

m = 9.1 ×10-31 Kg

\ Dv = 7.98 × 1012 m/sec

 

  1. n= = = 5.45 ´1017 s–1

½ mv2 = hn–hno

= (6.63´10–27 ergs–sec) (5.45´1017s–1) –2.62 ´10–9 ergs

            K.E. = 9.93 ´10–10 ergs

 

  1. Let the number of photons required = n

n

n =         =  = 27.6 = 28  photons

 

  1. Energy corresponding to the incident photon

E = 6.626 × 10-34 × 3.2 × 1016

= 2.12 ×10-17 J

K.E of photoelectron = th energy of photon

= ×2.12×10-17

= 1.590 × 10-17 J

hn = hn0 + KE

2.12 × 10-17 = hv0 + 1.590 ×10-17

hv0 = 5.297 × 10-18 J

n0 =

= 7.995 1015 Hz

  1. Time taken for one complete revolution is given by T =

T a n3

 

, n2 = 4  Þ n = 2

\ the ratio of time taken = 13 : 23

= 1:8

 

  1. For nth orbit mvr =

= mv =

Again from de-Beoglie’s equation we have

 

\

 

l =

 

  1. Energy of a single photon = h n =

= = 4.42 ´10–19J                                      …(1)

Energy emitted by the bulb = ´150 J = 12 J            …(2)

Let ‘n’ photons are emitted per second

n´4.42´10–19J = 12 J

            n = 27.2 ´1018

 

  1. According to Rydberg equation

= RZ2

for lowest frequency in Lyman series n2 = 2 and n1 = 1

\ = 1.09678 ´107 ´  (Z = 1 for hydrogen)=

l= 1.2156´10–7 metre

            l = 1215.6 Å

Frequency = n = =

            = 2.4679´1015Hz

Energy E = hn = 6.626´10–34 ´2.4679´1015

= 16.352´10–19 J = eV = 10.2 eV

Now energy for corresponding spectral line of Li2+ species

= Z2EH (for Li2+, Z =3)

= (3)2 ´10.2 = 91.80 eV

 

 

Level – II
  1. Energy corresponding to the photon =

= 3.822 ×10-19 J

the power output = 2.79 Watt/m2

= 2.79 J/sec/m2

\no. of photons emitted per sq. meter sec =

= 7.30 ×1018

 

  1. i) KE of an electron in the orbits of hydrogen and hydrogen like atoms

=

= 3.4 eV

  1. ii) =

= 6.654 × 10-10 m

= 6.654 Ao

 

  1. a) Energy of photon , E = W + KE

= 1.82 + 0.73 = 2.55 eV

  1. b) The energy of an electron in nth orbit of hydrogen atom is given as

En =

E1 = –13.6 eV

E2 =

E3 = –1.51eV

E4 = –0. 85 eV

Obviously, the transition 2 ¬ 4 leads to the release of photon of energy

2.55 eV as

QDE = | E2 – E4| = 2.55 eV

 

  1. If lo be the threshold wavelength and f be the work function
    lo = =  .

For copper = lo =  = 276 nm

For sodium = lo =  =540 nm

For caesium = lo =  = 654 nm

  1. a) =

= 1.46 × 10-9 m

  1. b)

when velocity is same l µ

 

 

 

 

  1. a) Energy corresponding to 8205.8 Ao =

= 2.422 × 10-19 J

= 1.572 eV

 

1.512 eV = E1H × (1)2 ×

1.512 eV =

E1 H = 13.608 eV

\ Ionisation energy of hydrogen atom = 13.6 eV

 

  1. b)

= 916 Ao

 

  1. Angular momentum of the electron = 5.2728 × 10-34 kgm2/sec

= 5.272 × 10-34

n1 = 4.99 » 5

Angular momentum of electron after transition = 3.1636 × 10-34 Kgm2/sec

= 3.1636 × 10-34

n2 = 2.999 » 3

\ the electron makes trasition from . 5 ® 3

n =  = c × Rh × Z2

n = 3 × 1010 × 109678 × 32 ×

= 2.1058 × 1015 Hz

 

  1. Energy corresponding to n = 2 =

= 5.425 × 10-12 erg

5.425 × 10-12

l = 3.66 × 10-5  cms

 

 

n2 = 4

next level is n = 5

velocity of electron in 5th level of  He+ =

=  = 87.39 ×106 cm/sec

velocity of the electron in 5th level : velocity of light = 1:344

 

  1. Lowest energy in u.v. region is

E = hc RH

                6.626 ×10-34 × 3 ×108 × 1.09 × 107 ×

= 1.625 × 10-18 J

This is energy corresponding to 1 atom of H

For 1 g atom the energy is

E = 1.625 × 10-18 × 6.023×1023 J

= 978.753 KJ mole-1

 

  1. eVo = –f f =  work function

f =   = 3.63 eV

 

K.E. = hn – f

=     = 2.57 eV

\ Stopping potential = 2.57 Volts           

 

  1. a) E = hn

n =  = 5.99 ´1014 sec–1

  1. b) Energy associated with photon of blue light with wave length 450 nm or 450 ´10–9 m

E = =

                  = 4.417 ´10–19 J/ Atom

                  Above energy is greater than threshold energy (i.e, 3.97 ´10–19J).

                  Hence electron will be ejected.

 

  1. Kinetic energy of electron = hn – hncritical

=  –

KE =  = 1.22 eV

(Since 1240 nm = 1eV)

 

  1. 14. Energy required to stop the ejection of electron is equal to kinetic energy i.e., (½ mv2)

Given K.E = 0.24 eV

l of incident photon = 253. 7 nm

hn = hno + K.E

Work function (hno) = hn – K.E = – ½ mv2

=   = 4.89 –  0.24

            = 4.65 eV

 

  1. Let the number of moles of I2 dissociated are n

Total energy = 1 kJ

= 1000 Joule

NA ® Avogadro’s no.

 

On solving above equation we get

n = 4.17 ´10–3 moles

 

 

level – iiI

 

  1. a) Q 1eV = 1.602 ´ 10–12 erg

Also DE = = E3 – E2 = RH.c.h.Z2

or 16.52 ´ 1.602 ´ 10–12  = 109678 × Z2 ×´ 3 ´ 1010 ´ 6.626 ´ 10–27

\ Z2 = 8.74        \ Z » 3

  1. b) DE = E4 – E3 = RH.c.h.Z2

= 109678 ´ 3 ´ 1010 ´ 6.626 ´ 10–27 ´ 32 ´  = 1.744 ´10–10  erg

               = 108.87 eV

 

  1. c)  = 109678 ´ 9

               l = 1.01 ´ 10–6 cm

 

  1. d) E. = mu2     = m

= = 1.962 ´ 10–10 erg

               = 122.4 eV

 

  1. Let wavelength for the transition 5®4 and  be that of 4 ® 3 transition

=  = 4.502 × 10-7 m

= 450.2 mm

 

= 2.089 × 10-7 m = 208.9 nm

shorter wavelength is , longer wavelength is

stopping potential for the photo electrons ejected by the light of wavelength

l2 is 3.95 Volts

Energy corresponding to l2 is

= 5.95 eV

since stopping potential = 3.95 V;

K.E. = 3.95 eV

E = f + KE1

5.95 = f + 3.95

f =  2.00 eV

Energy corresponding to l1 is

= 2.75 eV

E1 = f + K.E2

2.75 = 2.00 + KE2

KE2 = 0.75 eV

\stopping potential = 0.75 V

 

  1. Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell) = 10.20 + 17.0 = 27.20 eV = 27.20 ´ 1.602 ´ 10–12 erg

Q

\ 27.20 ´ 1.602 ´ 10–12 = RH ´ Z2 ´ h ´ c         … (1)

Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e., 3rd shell) = 4.25 + 5.95 = 10.20 eV

= 10.20 ´ 1.602 ´ 10–12 erg

\ 10.20 ´ 1.602 ´ 10–12 = RH ´ Z2 ´ h ´ c         … (2)

Dividing Eq. (1) by Eq. (2)

         n = 6

On substituting the value of n in Eqs. (1) or (2).

Z= 3

  1. Radius of 1st orbit of the He+ ion =

= 0.265  = 0.265 × 10-10 m

Uncertainty in the location = 1 % of the radius of the 1st orbit

= 0.265 × 10-12 m

Dp × Dx ³

Dv ³

³

³ 2.18 × 108 m/sec

velocity of the electron

v =  m/sec

 

= 4.368 ×106 m/sec (since Z = 2, n = 1)

 

  1. a) I.P. of the atom is

I.P. = E¥ – E1

= 0 – (-15.6) = 15.6 eV

  1. b) Series limit at n = 2 is

E = E¥ – E2

= 0-(-5.3) = 5.3 eV

= 5.3 eV

= 5.3 eV

l = 233.9 nm

  1. c) D E for the transition n = 3 ® n = 1 is

D E = E3 – E1

= -3.08 – (-15.6)

= 12.52 eV

= 2.005 ×10-18 J

= 2.005 ×10-18

 = 1.009 × 107 m-1

  1. d) The difference in energy between n = 1 and n = 2 is 10.3 eV
  2. i) Since the incoming electron has 6 eV energy, it will not be able to do anything hence it will not interact with the atom.
  3. ii) In case it has 11 eV energy, it will impact 10.3 eV energy to this atom and (11 eV – 10.3 eV) 0.7 eV energy is carried by this electron. Due to transfer of 10.3 eV of energy it will jump from n = 1 to n = 2 level.

 

 

  1. 1 g H contains          = N atoms

\ 1.8 g contains      = N ´ 1.8 atoms

= 6.023 ´ 1023 ´ 1.8

= 10.84 ´ 1023 atoms

 

(a) \ No. of atoms in III shell  =

      = 292.68 ´ 1021 atoms

\ No. of atoms in II shell =

               = 162.6 ´ 1021 atoms

and No. of atoms in I shell =

               = 628.72 ´ 1021 atoms

 

(b)  When all the atoms return to I shell, then

E¢ = (E3 – E1) ´ 292.68 ´ 1021

=  ´ 1.602 ´ 10–19 ´ 292.68 ´ 1021

= 5.668 ´ 105 joule

E¢¢ = (E2 – E1) ´ 162.6 ´ 1021

=  ´ 1.602 ´ 10–19 ´ 162.6 ´ 1021

= 2.657 ´ 105 joule

E = E¢ + E¢¢ = 5.668 ´ 105  + 2.657 ´ 105 joule

= 832.50 kJ

 

  1. Energy of photon = kinetic energy of the photo electron and threshold preparation.

hn­1 = KE1 + hn0                               …(1)

hn2 = KE2 + hn0                               …(2)

Multiplying (1) by 2 and substracting equation (2) from it

2hn1 – hn2 = hn0 (Q2KE1 = KE2)

2n1 – n2 = n0

n0 =  (1Å = 10–10 m)

n0 = 1.1483 ´ 1015 sec–1

Also l =  = 2.6126 ´ 10–7 m

         = 2612.6Å

 

  1. Energy of I orbit of H like atom = 4Rh

= 4 ´ 2.18 ´ 10–18 joule

E1  for H = –2.18 ´ 10–18 J

Q EH like atom = E1H ´ Z2

\ –4 ´ 2.18 ´ 10–18 = –2.18 ´ 10–18 ´ Z2

\ Z = 2

i.e., Atomic no. of H like atom is 2 or it is He+ ion.

 

  1. a) For de-excitation of  electron  in He+ from n2 = 2 to n1 = 1

E2 – E1 =

Now E1 = 4Rh

\ E2 = –  = –Rh

\ E2 – E1 = 3Rh = 3 ´ 2.18 ´ 10–18 J

\ E2 – E1 =

\ l =  = 303.89 ´ 10–10 m

               = 303.89 Å

  1. b) Radius (r1) of H like atom =  = 2.645 ´ 10–9 cm

 

 

  1. Energy given to I2 molecule

            =  =  = 4.417 ´ 10–19 J

Also energy used for breaking up to I2 molecule  =  = 3.984´10–19 J

\ Energy used in imparting kinetic energy to two I atoms

= [4.417 – 3.984] ´ 10–19 J

\ K.E./ iodine atoms  = [(4.417 – 3.984)/2] ´ 10–19

= 0.216 ´ 10–19 J

 

  1. Wave length emitted is in UV region and thus n1 = 1; For H atom

\

= 1.097 ´ 107

\ n = 2

Also the energy released is due to collision and all the kinetic energy is released in form of photon. Thus

=

or  ´ 1.67 ´ 10–27 ´ u2 =

\ u = 4.43 ´ 104 m sec–1

 

  1. Let x and y be the amounts of F and Cl atoms in the given mixture. Since the ionisation of atoms absorbs energy, we can write.

x(6.023 ´ 1023 mol–1) (27.91 ´ 10–22 kJ) + y (6.023 ´ 1023 mol–1) (20.77 ´ 10–22 kJ)

= 272.2 kJ                …(1)

Since the addition of electron to atoms releases energy, we can write

x(6.023 ´ 1023 mol–1) (5.53 ´ 10–22 kJ) + y(6.023 ´ 1023 mol–1) (5.78 ´ 10–22 kJ)

= 68.4 kJ                              …(2)

Solving for x and y, we get

x = 0.054 mol and y = 0.144 mol.

Percentage of F atoms  =  ´ 100 = 27.27

Percentage of Cl atoms = 100 – 27.27  = 72.73

 

  1. The basic expressions in Bohr model of the atom are as follows.

(a)  Stability of the circular motion of the electron, i.e.

Attractive force  = Centrifugal force

…(1)

Quantization of angular momentum

mvr =                                                     …(2)

Eliminating v in the above two expressions, we get

=

This gives r=  n2 …(3)

  1. b) Now for the given problem

m = (200)me and Z = 3

Hence,

r = n2

= n2(8.856 ´ 10–14 m)

Equation Eq. (3) with the first Bohr orbit for the hydrogen atom, we get

n2

or n2 = 200 ´ 3  or n =  = 24

 

(c)  The energy of the electron in the Bohr model of atom is

E = KE + PE = mn2

Using Eq. (I) , we get

E =

Substituting the expression of r from Eq. (3), we get

E =

=

= (3.899 ´ 10–15 J)

Hence, for the transition 1 ¬ 3, we get

DE = (3.899 ´ 10–15J)  = 3.465 ´ 10–15 J

l =

= 5.736 ´ 10–11 m = 57.36 pm

 

 

  1. O2 ¾® ON +

O2 ¾® ON + ON

E = 498 ´103 J / mole

= per molecule = 8.268 ´10–19 J

Energy required for excitation = 1.967 eV

= 3.146 ´10–19J

Total energy required for photochemical dissociation of O2
= 8.268 ´10–19 + 3.146 ´10–19   = 11.414 ´ 10–19 J

= 11.414 ´10–19 J

l =

= 1.7415 ´10–7 m = 1741.5 Å                       

  1. Energy absorbed =

=

= 5.52 ´ 10–11 erg

= 5.52 ´ 10–18 joule

Now this energy is used in overcoming forces of attraction between surface of metal and imparting velocity to electron, therefore,

Eabsorbed = E used in attractive forces  + Kinetic energy of electron

\ Kinetic energy = 5.52 ´ 10–18 – 7.52 ´ 10–19 joule = 47.68 ´ 10–19 joule

 

  1. Energy of light absorbed by one photon =

Let n1  photons are absorbed, therefore,

Total energy absorbed  =

Now E of light re-emitted out by one photon =

Let n2 photons are re-emitted then,

Total energy re-emitted out  = n­2 ´

As given Eabsorbed ´  = Ere-emitted out

´ n1 ´   = n´

            \  =  ´  =

\  = 1.25

 

 

    Solutions to Objective Problems

Level – I

 

  1. Since average mass is M + 0.5

 

= M + 0.5

\ (B)

 

  1. For the last electron of Ga the configuration is 4p1. The added electron enters the vacant p orbital making configuration 4p2

For 4p electron n = 4 and l = 1

\ (A)

  1. We know that rn = 0.529  Å, = 0.529 (32–22) Å

(For hydrogen Z = 1, n = 1)

= 0 .529 ´5´10–8cm

= 2.645 ´10–8 cm

            \ (A)

 

  1. Total spin (s) of an atom is given by

s = n      (Where n is number of unpaired electrons)  =  ´ 4  = 2

            \(C)

 

  1. Orbital angular momentum =

for d, l  = 2
orbital angular  momentum  =

=

            \ (A)

 

  1. Hint: Isosters have same number of electrons and atoms

 

  1. Hint: 7+3×8+1 = 32

            \ (B)

 

  1. 12. For one molecule

E = hn

For 1.5 moles

E = 1.5 ´  6.023 ´1023 ´hn

= 1.5 ´ 6.023 ´1023 ´6.626 ´10–34 ´ 7.5 ´1014   = 4.48 ´105 J

\ (C)

  1. Hint: n1 = 2 and n2 = 3, 4, 5, …… corresponds to Balmer series
  2. Wave number of a spectral line

Energy =

= 6.626 ´10–34 ´3´108´5´105

                        = 9.93 ´10–23kJ

            \ (B)

 

 

  1. = RH

= 109678

= 109678 ´

l = = 1215.6 Å

\ (A)

  1. Hint: r µ
  2. Hint: En of H- like species =

 

 

Level – II

  1. Hint: For lowest wave length, the energy difference between two energy states should be highest.

 

  1. Energy difference between 3rd and 4th I.P.’s is very high.

 

 

  1. Number of quanta absorbed =

=

= = 1.065 ´1022

            \ (A)

 

  1. = RH

RH =

 

 

l =

= 486.0 nm

            \(C)

 

  1. One of the possible excited state of He+ ion should have energy

En =  eV

\ En =  = -6.04 eV

 

  1. Energy difference between 3rd and 4th I.P.’s is very high.

 

  1. The ion has configuration [Ar] 4s03d6

Before losing the electrons it would have been [Ar] 4s23d7

\ the ion is Co3+

\ (C)

  1. En of H- like species =

D E for H like species = D E for H × Z2

For Be3+ Z = 4

D E = 10.2 ×16 = 163.2

            \ (D)

 

  1. Hint: Isoelectronic species have same number of electrons.

 

  1. Hint: The order of energies of orbitals is

2s < 2p < 3p < 4s < 3d

The transition from higher to lower levels result in emission of light.

 

  1. … (i)

                                      …(ii)

Dividing equation (I) by equation (ii), we get

, ,    \ Z = 2

\ He+  \ (A)

 

  1. l = h /mv

=  = 1.060 ´ 10–36 m

            \ (B)

 

  1. =  R

=  (1.09678 ´ 107) ´ 22 ´ ,  = 4.3871 ´ 107

or l = 2.279 ´ 10–8 m

\(A)

 

  1. l = =

10–10 =

On solving we get

V  = 0.0826 Volt

            \(C)

  1. hc=

 = 1.6203 ×106 m-1

\ (B)

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