# Hints for Subjective Problems

#### LEVEL – I

1. b) Relate K.E. to λ with the help of an equation which relates λ of a particle to its mass and velocity.
2. Total energy emitted = no. of photons emitted × energy of one photon.

#### LEVEL – II

1. c) Energy required to stop ejection of electrons = K.E. of electrons.

#### LEVEL – III

1. Get the value of Rydberg constant for positronium species.

1. Energy of incident photon will be partially utilized in breaking the I2 molecule and residual energy in the form of K.E.
2. 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.

# Solution to Subjective Problems

#### LEVEL – I

1. a) λ = = = 0.03313 × 10–31 m

or 3.313 × 10–33 m

###### b)We are given,

Kinetic energy, mv2 = 4.55 ×10–25J

m = 9.1 ×10–31 kg

h = 6.6 ×10–34 kg m2s–1

= 6.6 ×10–34 J–sec

½ ×(9.1×10–31) v2 = 4.55 ×10–25

v2 = = 106

or v = 103 m sec–1

λ = h/mv =

λ = 7.25 ×10–7 m.

1. c) mv2= 100 eV ×602 × 10–19

= 160.2 × 1019J

or

= 160.2 × 10–19 kg m2s–2

V =

= = 59.3 × 105 m/s

Now λ =

= = 0.1227 nm

or

1.227Å

1. i) Energy of the electron

En =

= – = – 2.43 × 10–12 erg

1. ii) Radius of the III orbit

rn =

= 4.75 × 10–8 cm = 4.75 Å

iii) Frequency and wavelength

E3 = and

ΔE = E3 – E2 =

= = 19.44 × 10–12 erg

ΔE = hν

Therefore v (frequency of radiation emitted)

= = = 2.95 × 1015 Hz

C = νλ

λ = = 1.017 × 10–5 = 1017 × 10–8 cm

= 1017Å

3 Radius of the nth  orbit is given by

rn = for hydrogen Z = 1

r1 = = 0.53 × 10–8 cm = 0.53 Å

Also, radius of the nth orbit is related to the I orbit by the relation

rn  = n2r1

Therefore r2 = 4 × 0.53 = 2.12 Å

& r3  = 9 × 0.53 = 4.77 Å

Corresponding values of He+(Z = 2) are

r1 = = 0.265 Å r2 = = 1.06Å

r3 = = 2.385 Å

1. I.E. = eV per atom

= eV = 0.85 eV

or 1.36 × 10–19 J per atom

4.18 J = 1 cal

I.E. = cal per atom

= 0.325 × 10–19 cal per atom

or 0.325 ×10–19 × 6.02 × 1023 cal/mole

= 19.5 kcal/mole

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

ΔE = Ef – Ei =

Hence ni = 1, nf = 3, E1 = –13.6 eV

ΔE = –13.6 = 12.08 eV.

1. Z = 1 for hydrogen atom

For Paschen series,  n1 = 3

For the first line, n2 = 4

= RZ2

= 109673 ×12 [0.0486] = 5331 cm–1

λ = = 1.875 ×10–4 cm.

1. =

λ = = 1215× 10-8 cm = 1215Å

λ = = 6565Å

8. The energy of each photon =

λ =

λ = 6630 nm

1. a) Energy required to remove the electron from n = 2 to n2 =

ΔE = 21.7 × 10–12 = 5.425 × 10–12 erg

Now, from Planck’s constant

E = hν =

λ =

= 3.67 × 10–5 cm

1. b) Ionisation energy for hydrogen

= –E1 (Energy of first Bohr orbit)

I.E = –E1 = – = = +13.6eV

Ionisation energy of Li2+

Z = 3

n = 2 ( for 1st exited state )

I.E = –

= 30.6 eV

Ionisation energy  of Be3+

Z = 4

n = 2 (for 1st exited state)

I.E =

= 54.4 eV

1. Energy of a single photon = h ν =

= = 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

LEVEL – II

1. a) v =

=

= 8.09 × 107 m/s

K.E. = mv2 = × 9.1 × 10–31 kg × (8.09 × 107 m/s)2

= 2.98 × 10–15J

or 1.86 × 104 eV

or 18.6 KeV

b)

= = 4.85 × 10–11 m

1. i) Li2+ is a H-like (1s1) particle, whose energy levels are given by

Ea = [for Li, Z (atomic weight) = 3]

=

The energy required to excite the electron from the I to the III orbit is

E3 – E1 = = 108.8 × 1.6 × 10–19 J

Q 1 eV = 1.6 × 10–19 J

Therefore, λ =

= 113.7 × 10–10 m = 113.7Å

1. ii) Three spectral lines will be observed in the emission spectrum of the above excited system corresponding to the transition shown below.

3 1 3 2 2 1

1. of spectral lines = = 3
2. a) i) Energy of photon , E = W + KE

= 1.82 + 0.73 = 2.55 eV

1. ii) 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

QΔE = | E2 – E4| = 2.55 eV

1. b) If λo be the threshold wavelength and φ be the work function
λo = .

For copper = λo = = 276 nm

For sodium = λo = =540 nm

For caesium = λo = = 654 nm

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

Given K.E = 0.24 eV

λ of incident photon = 253. 7 nm

hν = hνo + K.E

Work function (hνo) = hν – K.E = – ½ mv2

= 4.89 –  0.24  = 4.65 eV

1. λ =

= ∴ K.E. = =

= 9.65 × 10–23J or 6.03 × 10–4 eV

1. E3 for H = –2.41 ×10–12 er g

E2 for H = –5.42 ×10–12 erg

For a jump from 3rd to 2nd shell

ΔE = E3–E2 =

λ =

= = 6602.9 ×10–8 cm = 6603 Å

6.

5.2772 × 10–25 kg m sec–1

ΔVx = 5.79 × 105 ms–1

1. eVo = φ φ =  work function

φ =   = 3.63 eV

K.E. = hνφ

= = 2.57 eV

Stopping potential = 2.57 Volts

1. a) E = hν

ν =

= 5.99 ×1014 sec–1

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

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. Number of waves made by Bohr electron in an orbit is given by

n = rn = radius of nth orbit

vn = velocity of electron in nth the orbit

= = 2.98 3

Number of revolutions per second made by the electron =

= = 0.0243 × 1016

= 2.43 × 1014 revolutions per second

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

Total energy = 1 kJ

× n × NA = 1000 Joule

= 1000

On solving above equation we get

N = 4.17 × 10–3 moles

LEVEL – III

1. p = mv

and according to de-Broglie λ =

or p =

Let uncertainty in wavelength is Δλ which creates ΔP uncertainty in momentum.

p = p – Δp =

(as p decreases with increase in λ)

On substraction

Δp =

∴Δx =

Now since in 1 m wavelength uncertainty is = 10–6m

5 × 10–7 m wavelength uncertainty will be = 5 × 10–13 m

i.e. Δλ = 5 × 10–13m

On calculation Δx = 4 ×10–2 m

1. i) Q 1eV = 1.602 × 10–12 erg

Also ΔE = = 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. ii) ΔE = E4 – E3 = RH.c.h.Z2

= 109678 × 3 × 1010 × 6.626 × 10–27 × 32 × = 1.744 ×10–10  erg = 108.87 eV

iii) = 109678 × 9

λ = 1.01 × 10–6 cm

1. iv) E. = mu2 = m

= = 1.962 × 10–10 erg

= 122.4 eV

1. μ = me [for hydrogen atom]

for postronium species μ =

Q I.E. R

I.E. for e+   = × I.E. of hydrogen  =  × 13.6 = 6.8 eV

1. Number of emitted types of photons =

In initial excited state electrons are emitting three different types of photons

= 3

or n = 3

i.e. when an electron is in n = 3 it can emit 3 different types of photons. Now, as according to the question initially electrons were found in two different excited state, the other excited state is obviously n = 2

Now, when sample is exposed to 2.88 eV the resulting electrons emit 10 different type of photons some having energy equal to 13.056 eV or less than 13.056 eV

or = 10

n2 – n = 20

or n = 5

When sample is exposed to 2.88 eV energy radiation electrons in n = 2 move to
n = 5 (as energy difference in n = 2 and n = 5 is 2.88 eV).

After excitation some electrons are in n = 5, some in n = 3 and some may be in n = 2 all of them when come back to lower excited state emit photons of different energies maximum value is 13.05 eV when electrons come back from n = 5 to n = 1.

Initially excited electrons are either inn = 3 or n = 2, the maximum energy difference is between n = 3 and n = 1 and minimum energy difference is n = 3 and n = 2.

∴ΔEmax= 13.6

=

= 13.6 – 1.51 = 12.09 eV

ΔEmin=

= = 3.4 – 1.5 = 1.9 eV

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

hν1 = KE1 + hν0 …(1)

hν2 = KE2 + hν0 …(2)

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

2hν1 – hν2 = hν0 (Q2KE1 = KE2)

2ν1ν2 = ν0

ν0 = (1Å = 10–10 m)

ν0 = 1.1483 × 1015 sec–1

Also λ = = 2.6126 × 10–7 m  = 2612.6Å

1. 1 g H contains = N atoms

1.8 g contains = N × 1.8 atoms

= 6.023 × 1023 × 1.8

= 10.84 × 1023 atoms

1. a) 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

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. E1of atoms = – 13.6 eV

Energy give to H atom =

= 1.933 × 10–18J = 12.07 eV

Energy of H atom after excitation = – 13.6 + 12.07 = – 1.53 eV

Q = n2

n2 = = 9

n = 3

Thus electron in H atoms is excited to 3rd shell

I induced λ1 =

Q E1 = – 13.6 eV; E3 = – 1.53 eV

λ1 = = 1028 × 10–10 m

λ = 1028 A

II induced λ2 =

Q E1= – 13.6 eV; E2 = eV

λ = 1216 × 10–10m = 1215Å

III induced λ3 =

Q E1 = – 13.6 eV; E2 = l; E3 = eV

∴λ3 = = 6568 × 10–10 = 6558Å

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 =

λ = = 303.89 × 10–10 = 303.89 Å

1. b) Radius (r1) of H like atom = = 645 × 10–9 cm
2. 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.
2. 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

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

E = KE + PE = mν2

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

ΔE = (3.899 × 10–15J) = 3.465 × 10–15 J

λ =

= 5.736 × 10–11 m = 57.36 pm

1. E of light absorbed in one photon =

Let n1 photons are absorbed, therefore,

Total energy absorbed =

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

Let n2 photons re-emitted out = n2 ×

At given Eabsorbed × × Ere-emitted out

= 0.59

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

# LEVEL – I

1. En

= 1 :1

(B)

1. Bohr model can not explain the spectrum of any species having multi electrons. It can explain only the spectrum of hydrogen and hydrogen like species having single electron only.

(B)

1. The order of penetration power of s, p, d and f electrons are in order
s > p > d > f

(A)

1. Longest wavelength in Lyman Series of hydrogen atom arises from transition between n = 2 to n = 1 whose number is given by

=

Shortest wavelength in Balmer series of He+ arises from transition between
n = α to n = 2 whose wave number is given by

= R

=

(A)

1. According to Bohr theory

mvr =

According to de-Broglie

λ =

λ =

where r is the radius of the orbit having principal quantum number = n

Q (r n2) (here a3 = radius of 3rd orbit)

λ = = 6πa1

(B)

1. ΔxΔp =

If Δx = Δp

Δp2 =

(mΔv)2 =

Δv =

(A)

1. Colour in an ionic species arises when wavelength of excited photon lies in visible region. In all other cases except Cu2+(aq) energy required for excitement of electron from lower to higher level does not lie in visible region. But in Cu2+(aq) splitting of d-orbitals energy takes place under the influence of magnetic field of water and electron on excitement absorbs wavelength in visible region.

(C)

1. Total values of m or orientation in external magnetic field or orbitals are given
by 2l + 1

(C)

1. Orbital angular momentum =

for d, l  = 2
orbital angular  momentum  =

=

(A)

1. Experiment cannot be done on a single atom. Even a small weight of sample contains million of atoms.

(B)

1. rn= , r0α n2

r3 = 9r1 = 9x

Also mvrn =

mvr3 =

mv =

λ= = 6πx,

(B)

1. 12. For one molecule

E = hν

For 1.5 moles

E = 1.5 ×  6.023 ×1023 ×hν

= 1.5 × 6.023 ×1023 ×6.626 ×10–34 × 7.5 ×1014

= 4.48 ×105 J

(C)

1. The maximum number of possible sub-shells associated with a particular shell is equal to principal quantum number i.e. n 5

(B)

1. Wave number of a spectral line

Energy =

= 6.626 ×10–34 ×3×108×5×105

= 9.93 ×10–23kJ

(B)

1. = RH

= 109678

= 109678 ×

λ = = 1215.6 Å

(A)

# LEVEL – II

1. It is orbital not orbit [orbit was proposed by Bohr] which is a three dimensional area where probability of finding electron is maximum. Orbital word was coined by quantum mechanics.

(A)

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

ΔE1 2 = = 10.2 eV

ΔE2 3 = = 1.9eV

On moving  from lower value of n to higher value of n energy difference between successive energy levels decreases.

(A)

1. = RH

RH =

λ =

= 486.0 nm

(C)

1. Number of orientation of orbitals in space in presence of external magnetic field is given by 2l + 1. Since value of l for sub-shell is 3, possible orientations will be 7.

(C)

1. Value of l are given by 0 to n – 1. For an electron in n = 3, value of l can be 0, 1 or 2.

(D)

1. Energy difference between 3rd and 4th I.P.’s is very high.
2. ΔE = hv = = = 3 × 10–17J

(A)

1. En = , where Z = atomic number  = 4

N = number of orbits

Be3+ (n = 2) (first exited state)

E1 = –13.6 eV, E2 = –13.6 × 16/4

(A)

1. Isoelectronic species have same number of electrons.
2. For a given value of l, value of m can be – l to + l. If l = 0, m can not have –1 value.

(B)

1. Electronic configuration of ultimate and penultimate shell in ground state of these elements is as follows:

Magnetic moment μ = Bohr magneton where N is number of unpaired electrons, μ will be highest for Cr having 6 unpaired electrons.

(B)

1. Vn =

V3 = or

(A)

1. … (i)

…(ii)

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

, Z = 2

He+  (A)

1. λ = h /mv = = 1.060 × 10–36

(B)

6