NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 2
Chapter Name : Structure Of Atom

Q2.1 (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Answer. (i) Mass of one electron = 9.10939 × 10 −31
kg
−31
∴ Number of electrons that weigh 9.10939 × 10 kg = 1

Number of electrons that will high 1g = (1 × 10
−3
kg)
1 −3
= × (1 × 10 kg)
−31
9.10939×10 kg

−3+31
= 0.1098 × 10
28
= 0.1098 × 10
27
= 1.098 × 10

(ii) Mass of one electron = 9.10939 × 10 kg
−31

Mass of one mole of electron = 9.10939 × 10 kg −31

Mass of one mole of electron = (6.022 × 10 ) × (9.10939 × 10
23 −31
kg)
−7
= 5.48 × 10 kg

Charge on one electron = 1.6022 × 10 coulomb
−19

Charge on one mole of electron = (1.6022 × 10 C) (6.022 × 10 −19 23
)
4
= 9.65 × 10 C

Page : 69 , Block Name : Exercise

Q2.2 (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of C. 14

( (Assume that mass of a neutron = 1.675 × 10 kg ). −27

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH at STP. 3

Will the answer change if the temperature and pressure are changed ?

Answer. (i) Number of electrons present in 1 molecule of methane (CH ) 4

{1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 × 10 23
molecules of methane
23 24
= 6.022 × 10 × 10 = 6.022 × 10

(ii) (a) NUmber of atoms of 14
C in 1 mole = 6.023 × 10 23

14
since 1 atom of C contains (14 − 6) i.e., 8 neutrons, the number of neutrons in 14g of

14 23 14 23
C is (6.023 × 10 ) × 8. Or, 14g of C contains (6.022 × 10 × 8) neutrons.

Number of neutrons in 7mg
23
6.022×10 ×8×7mg
=
1400mg

21
= 2.4092 × 10

(b) Mass of one neutron = 1.67493 × 10 −27
kg

Mass of total neutron in 7g of C 14

21 −27
= (2.4092 × 10 ) (1.67493 × 10 kg)

−6
= 4.0352 × 10 kg

(iii) (a) 1 mole of NH 3 = {1(14) + 3(1)}g of NH3

= 17g of NH3

23
= 6.022 × 10 molecules of NH3

Total number of protons present in 1 molecule of NH 3
23
= (6.023 × 10 ) (10)

24
= 6.023 × 10

Number of protons in 6.023 × 10 23
molecules of NH 3
23
= (6.023 × 10 ) (10)

24
= 6.023 × 10

protons
24
⇒ 17 g of NH3 contains (6.023 × 10 )

Number of protons in 34 mg of NH 3

24
6.022×10 ×34mg
=
17000mg

22
= 1.2046 × 10

(b) Mass of one proton = 1.67493 × 10 −27
kg

Total mass of protons in 34 mg of NH 3
−27 22
= (1.67493 × 10 kg) (1.2046 × 10 )

−5
= 2.0176 × 10 kg

The number of protons, electrons, and neutrons in an atom is independent of
temperature and pressure conditions. Hence, the obtained values will remain unchanged
if the temperature and pressure is changed.

Page : 69 , Block Name : Exercise

Q2.3 How many neutrons and protons are there in the following nuclei ?
56
, , , , .
13 16 24
C O Mg Fe 88 Sr
6 8 12
26 38

Answer. C:
13

6

Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = ( Atomic mass ) - (Atomic number)
= 13 − 6 = 7

:
16
O
8

Atomic mass = 16
Atomic number = 8

Number of protons = 8
Number of neutrons = (Atomic mass)-(Atomic number)
= 16 − 8 = 8
24

12
:
Mg

Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) - (Atomic number)
= 24 − 12 = 12

56
Fe :
26

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = ( Atomic mass ) − ( Atomic number )

= 56 − 26 = 30

88 Sr :
38

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass)-(Atomic number)

= 88 − 38 = 50

Page : 69 , Block Name : Exercise

Q2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass
(A).
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.

Answer. (i) 35

17
Cl

(ii)
233
U
92

(iii) Be
9

4

Page : 69 , Block Name : Exercise

Q2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the
frequency (ν) and wavenumber (ν ) of the yellow light.

Answer. From the expression,
c
λ =
v

We get,
c
v =
λ ………(1)

Where,
v = frequency of yellow light
8
c = velocity of light in vacuum = 3 × 10 m/s

−9
λ = wavelength of yellow light = 580nm = 580 × 10 m

Substituting the values in expression(i):
8
3×10 14 −1
v = = 5.17 × 10 s
−9
580×10

Thus, frequency of yellow light emitted from the sodium lamp
14 −1
= 5.17 × 10 s

Wave number of yellow light, v = ¯
¯¯ 1

λ
1 6 −1
= = 1.72 × 10 m
−9
580×10

Page : 69 , Block Name : Exercise

Q2.6 Find energy of each of the photons which
(i) correspond to light of frequency 3 × 10 Hz 15

(ii) have wavelength of 0.50 Å.

Answer. (i) Energy (E) of a photon is given by the expression,
E = hv

Where,
−34
h = Planck's constant = 6.626 × 10 Js
15
v = frequency of light = 3 × 10 Hz

Substituting the values in the given expression of E:
−34 15
E = (6.626 × 10 ) (3 × 10 )

−18
E = 1.988 × 10 J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,
hc
E =
λ
−34
h = Planck's constant = 6.626 × 10 Js
8
c = v velocity of light in vacuum = 3 × 10 m/s

Substituting the values in the given expression of E:
−34 8
(6.626×10 )(3×10 )
−15
E = −10
= 3.976 × 10 J
0.50×10

−15
∴ E = 3.98 × 10 J

Page : 69 , Block Name : Exercise

Q2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is
s.
−10
2.0 × 10

Answer. Frequency (v) of light = 1/period
1 9 −1
= = 5.0 × 10 s
−16
2.0×10 s

Wavelength (λ) of light =
c

v

Where,
8
c = velocity of light in vacuum = 3 × 10 m/s

Substituting the value in the given expression of λ :
8
3×10 −2
λ = 9
= 6.0 × 10 m
5.0×10

Wave number (v ) of light =
¯
¯¯ 1 1 1 −1
= −2
= 1.66 × 10 m = 16.66m
λ 6.0×10

Page : 69 , Block Name : Exercise

Q2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of
energy?

Answer. Energy (E) of a photon = hv
Energy (E ) of 'n' photons = nhv
n

En λ
⇒ n =
hc

Where,
−12
λ = wavelength of light = 4000pm = 4000 × 10 m
8
c = velocity of light in vacuum = 3 × 10 m/s
−34
h = Planck's constant = 6.626 × 10 Js

Substituting the values in the given expression of n:
−12
(1)×(4000×10 )
16
n = = 2.012 × 10
−34 3
(6.626×10 )(3×10 )

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10 .
16

Page : 69 , Block Name : Exercise

Q2.9 A photon of wavelength 4 × 10 m strikes on metal surface, the work function of the metal
−7

being 2.13eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission,
and (iii) the velocity of the photoelectron (1eV = 1.6020 × 10 J).
−19

Answer. (i) Energy (E) of a photon = hv =
hc

λ

Where,
−34
h = Planck's constant = 6.626 × 10 Js
∘
c = velocity of light in vacuum = 3 × 10 m/s

−7
λ = wavelength of photon = 4 × 10 m

Substituting the values in the given expression of E:
−34 8
(6.626×10 )(3×10 )
−19
E = = 4.9695 × 10 J
−7
4×10

Hence, the energy of the photon is 4.97 × 10 −19
J

(ii) The kinetic energy of emission E is given by k

= hv − hv0

= (E − W )eV
−19
4.9695×10
= ( −19
) cV − 2.13eV
1.6020×10

= (3.1020 − 2.13)eV

= 0.9720eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

1 2
mv = hv − hv0
2

2(hv−hv0 )
⇒ v = √
m

Where, (hv − hv ) is the kinetic energy of emission in joules and ‘m’ is the mass of the
0

photoelectron. Substituting the values in the given expression of v:
−10
2×(0.9720×1.6020×10 )J
v = √
−31
9.10939×10 kg

√ 12 2 −2
= 0.3418 × 10 m s

5 −1
v = 5.84 × 10 ms

Hence, the velocity of the photoelectron is 5.84 × 10 ms .
5 −1

Page : 69 , Block Name : Exercise

Q2.10 Electromagnetic radiation of wavelength 242 nm is just suf cient to ionise the sodium atom.
Calculate the ionisation energy of sodium in kJmol .
−1

Answer. Energy of sodium (E) =
NA hc

λ

23 −1 −34 8 −1
(6.023×10 mol )(6.626×10 Js)(3×10 ms )

=
−9
242×10 m
5 −1
= 4.947 × 10 Jmol

3 −1
= 494.7 × 10 Jmol

−1
= 494kJmol

Page : 69 , Block Name : Exercise

Q2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the
rate of emission of quanta per second.

Answer. Power of bulb, P = 25 \text { Watt }=25 \mathrm{Js}^{-1}
Energy of one photon, E = = hv =
hc

λ

Substituting the values in the given expression of E:
−34 8
(6.626×10 )(3×10 )
−20
E = = 34.87 × 10 J
−6
(0.57×10 )

−20
E = 34.87 × 10 J

Rate of emission of quanta per second
25 19 −1
= −20
= 7.169 × 10 s
34.87×10

Page : 69 , Block Name : Exercise

Q2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation
of wavelength 6800 Å. Calculate threshold frequency (v ) and work function (w ) of the metal. 0 0

Answer. Threshold wavelength of radian (λ ) = 6800A 0 =6800×10
−10
m

Threshold frequency (v ) of the metal 0

8 −1
c 3×10 ms 14 −1
= = −7
= 4.41 × 10 s
λ0 6.8×10 m

Thus, the threshold frequency (v ) of the metal is 4.41 × 10 0
14
s
−1

Hence, work function (W ) of the metal = hv o 0

−34 14 −1
= (6.626 × 10 Js) (4.41 × 10 s )

−19
= 2.922 × 10 J

Page : 69 , Block Name : Exercise

Q2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes
transition from an energy level with n = 4 to an energy level with n = 2?

Answer. The n = 4 and n
i f = 2 transition will give rise to a spectral line of the Balmer series. The
energy involved in the transition is given by the relation,
−18 1 1
E = 2.18 × 10 [ − 2
]
2
n n
i f

Substituting the values in the given expression of E:
−18 1 1
E = 2.18 × 10 [ 2
− 2
]
4 2

−18 1−4
= 2.18 × 10 [ ]
16

−18 3
= 2.18 × 10 × (− )
16

−19
E = − (4.0875 × 10 J)

The negative sign indicates the energy of emission.
Wavelength of light emitted (λ) =
hc

E

hc
(since E = )
λ

Substituting the values in the given expression of λ:
−34 8
(6.626×10 )(3×10 )

λ = −19
4.0875×10
−7
λ = 4.8631 × 10 m
−9
= 486.3 × 10 m

= 486nm

Page : 69 , Block Name : Exercise

Q2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?
Compare your answer with the ionization enthalpy of H atom ( energy required to remove the
electron from n =1 orbit).

Answer. The expression of energy is given by,
−18 2
−(2.18×10 )Z
En =
2
n

Where,
Z = atomic number of the atom

n = principal quantum number

For ionization from n 1 = = 5 to n 2 = ∞ ,

ΔE = E∞ − E5
−18 2 −18 2
−(2.18×10 J)(1) −(2.18×10 J)(1)
= [{ } − { }]
(∞)2 (5)2

−18 1 1
= (2.18 × 10 J) ( ) ( since = 0)
(5)2 ∞

−18
= 0.0872 × 10 J
−30
ΔE = 8.72 × 10 J
−20
Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10 J

Energy required for n1 = 1 to n = ∞

ΔE = E∞ − E1
−18 2 −18 2
−(2.18×10 )(1) −(2.18×10 )(1)
= [{ } − { }]
(∞)2 (1)2

−18
= (2.18 × 10 ) [1 − 0]

−18
= 2.18 × 10 J

Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as
compared to that in the ground state.

Page : 70 , Block Name : Exercise

Q2.15 What is the maximum number of emission lines when the excited electron of a H atom in n =
6 drops to the ground state?

Answer. When the excited electron of an H atom in n = 6 drops to the ground state, the following
transitions are possible:

Hence, a total number (5 + 4 + 3 + 2 + 1)15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the n th
level drops down to the ground
n(n−1)
state is given by
2

Given,
n=6
6(6−1)
Number of spectral lines = 2
= 15

Page : 70 , Block Name : Exercise

Q2.16 (i) The energy associated with the rst orbit in the hydrogen atom is
. What is the energy associated with the fth orbit?
−18 −1
−2.18 × 10 J atom

(ii) Calculate the radius of Bohr’s fth orbit for hydrogen atom.

Answer. (i) Energy associated with the fth orbit of hydrogen atom is calculated as:
−18
−(2.18×10 ) −18
−2.18×10
E3 = =
2 25
(5)

−20
E5 = −8.72 × 10 J

(ii) Radius of Bohr’s n th
for hydrogen atom is given by,
2
rn = (0.0529nm)n

For,
n = 5
2
r5 = (0.0529nm)(5)

r5 = 1.3225nm

Page : 70 , Block Name : Exercise

Q2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of
atomic hydrogen.

Answer. For the Balmer series, n = 2. Thus, the expression of wavenumber (v ) is given by,
i
¯
¯¯

¯
¯¯ 1 1 7 −1
v = [ − ] (1, 097 × 10 m )
(2)2 n
2
f

Wave number (v ) is inversely proportional to wavelength of transition. Hence, for the longest
¯
¯¯

wavelength transition, v has to be the smallest.¯
¯¯

¯
¯¯
For v to be minimum, nf should be minimum. For the Balmer series, a transition from ni

= 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:

¯
¯¯ 7 1 1
v = (1.097 × 10 ) [ − ]
2 2
2 3

¯
¯¯ 7 1 1
v = (1.097 × 10 ) [ − ]
4 9

9 − 4
7
= (1.097 × 10 ) ( )
36

5
7
= (1.097 × 10 ) ( )
36

¯
¯¯ 6 −1
v = 1.5236 × 10 m

Page : 70 , Block Name : Exercise

Q2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the
rst Bohr orbit to the fth Bohr orbit and what is the wavelength of the light emitted when the
electron returns to the ground state? The ground state electron energy is −2.18 × 10 ergs.
−11

Answer. Energy (E) on the n th
Bohr orbit of an atom is given by,
−18 2
−(2.18×10 )Z
En =
n2

W here,

Z = atomic number of the atom
−11
Ground state energy = −2.18 × 10 ergs

−11 −7
= −2.18 × 10 × 10 J

−18
= −2.18 × 10 J

Energy required to shift the electron from n = 1 to n = 5 is given as:
ΔE = E5 − E1
−18 2
−(2.18×10 )(1)
−18
= 2
− (−2.18 × 10 )
(5)

−18 1
= (2.18 × 10 ) [1 − ]
25

−18 24 −18
= (2.18 × 10 )( ) = 2.0928 × 10 J
25

Wavelength of emitted light =
hc

E
−34 8
(6.626 × 10 ) (3 × 10 )
=
−18
(2.0928 × 10 )

−8
= 9.498 × 10 m

Page : 70 , Block Name : Exercise

Q2.19 The electron energy in hydrogen atom is given by E . Calculate
−18 2
m = (−2.18 × 10 ) /n J

the energy required to remove an electron completely from the n = 2 orbit. What is the longest
wavelength of light in cm that can be used to cause this transition?

Answer. Given,
−18
2.18×10
En = − J
2
n

Energy required for ionization from n = 2 is given by,
ΔE = E∞ − E2
−18 −18
−2.18×10 −2.18×10
= [( ) − ( )] J
2 2
(∞) (2)

−18
2.18×10
= [ − 0] J
4

−18
= 0.545 × 10 J

−19
ΔE = 5.45 × 10 J
hc
λ =
ΔE

Here, λ is the longest wavelength causing the transition.
−34 8
(6.626×10 )(3×10 )
−7
λ = −19
= 3.647 × 10 m
5.45×10
−10
= 3647 × 10 m

= 3647 Å.

Page : 70 , Block Name : Exercise

Q2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 10 ms .
7 −1

Answer. According to de Broglie’s equation,
h
λ =
mv

Where,
λ = wavelength of moving particle

m = mass of particle

v = velocity of particle

h = Planck's constant

Substituting the values in the expression of λ :
−34
6.626×10 Js
λ = −31 7 −1
(9.10939×10 kg)(2.05×10 ms )

−11
λ = 3.548 × 10 m

Hence, the wavelength of the electron moving with a velocity of
m.
7 −1 −11
2.05 × 10 ms is 3.548 × 10

Page : 70 , Block Name : Exercise

Q2.21 The mass of an electron is 9.1 × 10 −31
kg . If its K.E. is 3.0 × 10
−25
J . calculate its
wavelength.

Answer. From de Broglie’s equation,
h
λ =
mv

Given,
Kinetic energy (K.E) of the electron = 3.0 × 10
−25
J

Since K.E = mv 1 2

2

2K.E
∴ Velocity (v) = √
m

−25
2(3.0×10 J)
= √ −31
9.10939×10 kg

4
= √6.5866 × 10
−1
v = 811.579ms

Substituting the value in the expression of λ :
−34
6.626 × 10 Js
λ =
−31 −1
(9.10939 × 10 kg) (811.579ms )

−7
λ = 8.9625 × 10 m

Hence, the wavelength of the electron is 8.9625 × 10 −7
m

Page : 70 , Block Name : Exercise

Q2.22 Which of the following are isoelectronic species i.e., those having the same number of
electrons?
+ + 2+ 2+ 2
Na ,K , Mg , Ca , s , Ar

Answer. Missing

Page : 70 , Block Name : Exercise

Q2.23 (i) Write the electronic con gurations of the following ions:
(a) H
−

(b) Na
+

(c) O 2−

(d) F
−

(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s 1

(b) 2p amd
3

(c) 3p ?5

(iii) Which atoms are indicated by the following con gurations ?
(a) [He]2s 1

(b) [Ne]3s 3p 2 3

(c) [Ar]4s 3d 2 1

Answer. (i) (a) H ion
−

The electronic con guration of H atom is 1s . 1

A negative charge on the species indicates the gain of an electron by it.
− 2
∴ Electronic configuration of H = 1s

(b) Na ion
+

The electronic con guration of Na atom is 1s 2s 2p 3s . 2 2 6 2

A positive charge on the species indicates the loss of an electron by it.
+ 2 2 6 0 2 2 6
∴ Electronic configuration of Na = 1s 2s 2p 3s or 1s 2s 2p

(c) O ion
2−

The electronic con guration of 0 atom is 1s 2s 2p 2 2 4

A negative charge on the species indicates that two electrons are gained by it.
2− 2 2 6
∴ Electronic configuration of O ion = 1s 2s p

(d) F ion
−

The electronic con guration of F atom is 1s 2s 2p 2 2 5

A negative charge on the species indicates the gain of an electron by it.
− 2 2 6
∴ Electron configuration of F ion = 1s 2s 2p

(ii) (a) 3s 1

Completing the electron con guration of the element as
2 2 6 1
1s 2s 2p 3s

∴ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 =11
∴Atomic number of the element = 11

(b) 2p 3

Completing the electron con guration of the element as
2 2 3
1s 2s 2p

∴Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
∴Atomic number of the element = 7

5
(c) 3p

Completing the electron configuration of the element as

2 2 5
1s 2s 2p .

∴Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
∴Atomic number of the element = 9
1
(iii) (a) [He] 2s

1 2 1
The electronic configuration of the element is [He]2s = 1s 2s .

∴ Atomic number of the element =3
Hence, the element with the electronic con guration [He]2s 1
is lithium (Li)
2 3
(b) [Ne] 3s 3p

2 3 2 2 6 2 3
The electronic configuration of the element is [Ne] 3s 3p = 1s 2s 2p 3s 3p

∴ Atomic number of the element = 15
Hence, the element with the electronic con guration [Ne]3s 3p is phosphorus (P).
2 3

(c) [Ar]4s 3d 2 2

The electronic con guration of the element is [Ar]4s 3d2 1 2 2
= 1s 2s 2p 3s 3p 4s 3d
6 2 6 2 1

Atomicnumberof theelement = 21

Hence, the element with the electronic con guration [Ar]4s 3d is scandium(Sc).
2 1

Page : 70 , Block Name : Exercise

Q2.24 What is the lowest value of n that allows g orbitals to exist?

Answer. For g-orbitals, L = 4.
As for any value 'n' of principal quantum number, the Azimuthal quantum number (l) can have a
value from zero to (n - 1).
∴For l = 4, minimum value of n = 5.

Page : 70 , Block Name : Exercise

Q2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and m for this l

electron.

Answer. For the 3d orbital:
Principal quantum number (r1) = 3
Azimuthal quantum number (l) =2
Magnetic quantum number (m1) = - 2, -1, 0, 1,2.

Page : 70 , Block Name : Exercise

Q2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of
protons and (ii) the electronic con guration of the element.

Answer. (i) For an atom to be neutral, the number of protons is equal to the number of electrons.
∴ Number of protons in the atom of the given element = 29

(ii) The electronic con guration of the atom is 1s 2s 2p 3s 3p 4s 3d
2 2 6 2 6 2 10

Page : 70 , Block Name : Exercise

Q2.27 Give the number of electrons in the species H , H and O
+ +

2 2 2

Answer. H :
+

2

Number of electrons present in hydrogen molecule (H ) = 1 + 1 = 2 2

+
∴ Number of electrons in H = 2 − 1 = 1
2

H2 :

Number of electrons in H2 = 1 + 1 = 2
+
O
2

Number of electrons present in oxygen molecule (O2 ) = 8 + 8 = 16

+
Number of electrons in O = 16 − 1 = 15
2

Page : 70 , Block Name : Exercise

Q2.28 (i) An atomic orbital has n = 3. What are the possible values of l and m ?
l

(ii) List the quantum numbers ( m and l ) of electrons for 3d orbital.
l

(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

(i)n = 3( Given )

For a given value of n, l can have values from 0 to (n − 1) .
Answer.
∴ For n = 3

l = 0, 1, 2

For a given value of l, m, can have (2l + 1) values.

For I = 0, m = 0

I = 1, m = −1, 0, 1

I = 2, m = −2, −1, 0, 1, 2

∴ For n = 3

I = 0, 1, 2

m0 = 0

m1 = −1, 0, 1

m2 = −2, −1, 0, 1, 2

(ii) For 3d orbital, l = 2

For a given value of l, m can have (2l + 1) values i.e., 5 values.

∴ For I = 2

m2 = −2, −1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.
For p-orbital, I = 1.
For a given value of n, l can have values from zero to (n — 1).
For l is equal to 1, the minimum value of n is 2.
Similarly,

For f-orbital, l = 4.
For I = 4, the minimum value of n is 5.
Hence, 1p and 3f do not exist.

Page : 70 , Block Name : Exercise

Q2.29 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1,
l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.

Answer. (a) n = 1, l = 0 (Given)
The orbital is 1s
(b) Fo n = 3 and I = 1
The orbital is 3p
(c) For n = 4 and l = 2
The orbital is 4d.
(d) For n = 4 and l = 3
The orbital 4f.

Page : 70 , Block Name : Exercise

Q2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = + ½
(b) n = 1, l = 0, ml = 0, ms = – ½
(c) n = 1, l = 1, ml = 0, ms = + ½
(d) n = 2, l = 1, ml = 0, ms = – ½
(e) n = 3, l = 3, ml = –3, ms = + ½
(f) n = 3, l = 1, ml = 0, ms = + ½

Answer. (a) The given set of quantum numbers is not possible because the value Of the principal
quantum number (n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible.
For a given value of n, 'l' car-I have values from zero to (n - 1)
For n =1, I = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) he given set of quantum numbers is not possible.
For n = 3,
l = 0 to (3 - 1)
l = 0 to 2 i.e., 0, 1, 2
(f) The given set of quantum numbers is possible.

Page : 70 , Block Name : Exercise

Q2.31 How many electrons in an atom may have the following quantum numbers?
(a) n = 4, m = −1/2
s

(b) n = 3, l = 0

Answer. (a) Total number of electron in an atom for a value of n = 2n 2

∴ For n = 4

2
Total number of electrons = 2(4)

= 32

The given element has a fully flled orbital as

2 2 6 2 6 2 10
1s 2s 2p 3s 3p 4s 3d

Hence, all the electrons are paired.
m, 1
∴ Number of electrons (having n = 4 and = − ) = 16
2

(b) n = 3, l = O indicates that the electrons are present in the 3s orbital. Therefore, the number of
electrons having n = 3 and I = O is 2.

Page : 71 , Block Name : Exercise

Q2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple
of the de Broglie wavelength associated with the electron revolving around the orbit.

Answer. Since a hydrogen atom has only one electron, according to Bohr's postulate, the angular
momentum of that electron is given by:
h
mvr = n , … … … (1)
2π

Where,
N = 1, 2, 3, …
According to de Broglie’s equation :
h
λ =
mv

h
or mv =
λ

Substituting the value of 'mv' from expression (2) in expression (1):
hr h
= n
^
λ 2π

or 2πr = nλ

Since '2nr' represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the
circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie's
Wavelength associated th the electron revolving around the orbit.

Page : 71 , Block Name : Exercise

Q2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer
transition n = 4 to n = 2 of He spectrum ?
+

Answer. For He ion, the wave number (v ) associated with the Balmer transition, n = 4 to n = 2 is
+ ¯
¯¯

given by :
¯
¯¯ 1 2 1 1
v = = RZ ( − )
2 2
λ n n
1 2

Where,

n1 = 2

n2 = 4

Z = atomic number of helium

¯
¯¯ 1 2 1 1
v = = R(2) ( − )
λ 4 16

4−1
= 4R ( )
16

¯
¯¯ 1 3R
v = =
λ 4

4
⇒ λ =
3R

According to the question, the desired transition for hydrogen will have the same wavelength as
that of He +

2 1 1 3R
⇒ R(1) [ − ] =
2 2
n n 4
1 2

1 1 3
[ − ] = … … … (1)
2 2
n n 4
1 2

By hit and trial method, the equality given by equation (1) is true only when
n1 = 1 and r2 = 2

∴ The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as
Balmer transition n = 4 to n = 2 of (He) spectrum.

Page : 71 , Block Name : Exercise

Q2.34 Calculate the energy required for the process.
+ 2+ −
He (g) → He (g) + e

The ionization energy for the H atom in the ground state is 2.18 × 10 −18
J atom
−1
.

Answer. Energy associated with hydrogen-like species is given by,
2
−18 Z
En = −2.18 × 10 ( )J
2
n

For ground state of hydrogen ator

ΔE = Ee − E1
2
−18 (1)
= 0 − [−2.18 × 10 { }] J
2
(1)

−18
ΔE = 2.18 × 10 J
+ 2+ −
He → He + e
(x) (x)

An electron is removed from n =1 to n =
ΔE = Em − E1
2
−18 (2)
= 0 − [−2.18 × 10 { 2
}]
(1)

−18
ΔE = 8.72 × 10 J

∴ The energy required for the process is 8.72x10 −18
J.

Page : 71 , Block Name : Exercise

Q2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which
can be placed side by side in a straight line across length of scale of length 20 cm long.

Answer.
1m = 100cm

−2
1cm = 10 m

Length of the scale = 20cm

−2
= 20 × 10 m

Diameter of a carbon atom = 0.15nm

−9
= 0.15 × 10 m
−9
One carbon atom occupies 0.15 × 10 m .

∴ Number of carbon atoms that can be placed in a straight line

−2
20×10 m
= −9
0.15×10 m

7
= 133.33 × 10

9
= 1.33 × 10

Page : 71 , Block Name : Exercise

Q2.36 2 × 10 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the
8

length of this arrangement is 2.4 cm.

Answer. Length of the given arrangement = 2.4 cm
Number of carbon atoms present = 2 x 10 8

∴ Diameter of carbon atom
−1
2.4 × 10 m
=
8
2 × 10
−16
=1.2 × 10 m

Diameter
∴ Radius of carbon atom =
2
−11
= 6.0 × 10 m

Page : 71 , Block Name : Exercise

Q2.37 The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b) number of
atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Answer.
Diameter
(a) Radius of zinc atom =
2

2.6A
=
2

−10
= 1.3 × 10 m

−12
= 130 × 10 m = 130pm

(b) Length of the arrangement = 1.6cm

−2
= 1.6 × 10 m

−10
Dlameter of zinc atom = 2.6 × 10 m

∴ Number of zinc atoms present in the arrangement

−10
1.6×10 m
= −10
2.6×10 m

8
= 0.6153 × 10 m

7
= 6.153 × 10

Page : 71 , Block Name : Exercise

Q2.38 A certain particle carries 2.5 × 10 of static electric charge. Calculate the number of
−16
C

electrons present in it.

Answer. Charge on one electron = 1.6022 x 10 −19
C
−19
⇒ 1.6022 × 10 C charge is carrled by 1 electron.

−16
∴ Number of electrons carrying a charge of 2.5 × 10 C
1 −16
= (2.5 × 10 C)
−19
1.6022×10 C

−19
= 1.6022 × 10 C

= 1560C

Page : 71 , Block Name : Exercise

Q2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining
X-rays. If the static electric charge on the oil drop is −1.282 × 10 C calculate the number of
−18

electrons present on it.

Answer.
−18
Charge on the oil drop = 1.282 × 10 C

−19
Charge on one electron = 1.6022 × 10 C

a Number of electrons present on the oil drop
−18
1.282×10 C
= −18
1.6022×10 C

−18
= 0.8001 × 10 C

= 8.0

Page : 71 , Block Name : Exercise

Q2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc.
have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium
etc. is used, what difference would be observed from the above results ?

Answer. A thin foil of lighter atoms will not give the same results as given with the foil of heavier
atoms.

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough
de ection of a-particles (positively charged).

Page : 71 , Block Name : Exercise

Q2.41 Symbols and can be written, whereas symbols and are not acceptable.
79 79 35 35
Br Br Br Br
35 79

Answer brie y.

Answer. The general convention of representing an element along with its atomic mass (A) and
atomic number (Z) is X. A
Z

74 35
Hence, Br is acceptable Br but is not acceptable.
35 79

79
can be written but Br cannot be written because the atomic number of an element is
Br
35

constant, but the atomic mass of an element depends upon the relative abundance of its isotopes.
Hence, it is necessary to mention the atomic mass of an element.

Page : 71 , Block Name : Exercise

Q2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons.
Assign the atomic symbol.

Answer. Let the number of protons in the element be x.
∴ Number of neutrons in the element

= x + 31.7% of x

= x + 0.317x

= 1.317x

According to the question,
Mass number of the element = 81
∴ (Number of protons + number of neutrons) = 81

⇒ x + 1.317x = 81

2.317x = 81

81
x =
2.317

= 34.95

∴x = 35

Hence, the number of protons in the element i.e., x is 35.
Since the atomic number of an atom is de ned as the number of protons present in its nucleus, the
atomic number of the given element is 35.
The atomic symbol of the element is 81 Br
35

Page : 71 , Block Name : Exercise

Q2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains
11.1% more neutrons than the electrons, nd the symbol of the ion.

Answer. Let the number of electrons in the ion carrying e negative charge be x. Then Number of
neutrons present
= x + 11.1% of x

= x + 0.111x

= 1.111x

Number of electrons in the neutral atom = (x - 1)
(When an ion carries a negative charge, it carries an extra electron)
∴Number of protons in the neutral atom = x - 1

Given, Mass number of the ion = 37
∴ (x − 1) + 1.111x = 37

2.111x = 38

x = 18

∘ −
The symbol of the ion is 17 Cl

Page : 71 , Block Name : Exercise

Q2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons
than electrons. Assign the symbol to this ion.

Answer. Let the number of electrons present in ion
Number of neutrons in it = x + 30.4% of x = 1.304 x
Since the ion is tripositive,
Number of electrons in neutral atom = x + 3
Number of protons in neutral atom = x + 3
Given, Mass number of the ion = 56
∴ (x + 3) + (1.304x) = 56

2.304x = 53

53
x =
2.304

x = 23

Number of protons = = x + 3 = 23 + 3 = 26
56 3+
∴ The symbol of the ion Fe
26

Page : 71 , Block Name : Exercise

Q2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from
microwave oven (b) amber light from traf c signal (c) radiation from FM radio (d) cosmic rays from
outer space and (e) X-rays.

Answer. The increasing order of frequency is as follows: Radiation from FM radio < amber light <
radiation from microwave oven < X cosmic rays.
The increasing order of wavelength is as follows: rays < Cosmic rays < X-rays < radiation from
microwave ovens < amber light < radiation of FM radio.

Page : 71 , Block Name : Exercise

Q2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons
emitted is 5.6 × 10 . calculate the power of this laser.
24

Answer. Power of laser = Energy with which it emits photons = (E =
M hc
)
λ

Power Where,
N = number of photons emitted
h = Planck's constant
c = velocity of radiation
λ =wavelength of radiation

Substituting the values in the given expression Of Energy (E) :
24 −34 4 −1
(5.6×10 )(6.626×10 Js)(3×10 ms )
E== −9
(337.1×10 m)

7
= 0.3302 × 10 J

6
= 3.33 × 10 J

6
Hence, the power of the laser is 3.33 × 10 J

Page : 71 , Block Name : Exercise

Q2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the
frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d)
number of quanta present if it produces 2 J of energy.

Answer. Wavelength of radiation emitted = 616 nm = 616 x 10 −9
m (Given)
(a) Frequency of emission (v)
c
v =
λ

where,
c = velocity of radiation
λ= wavelength of radiation

Substituting the values in the given expression of
8
3.0×10 m/s
v = −9
616×10 m

8 9 −3 −1
= 4.87 × 10 × 10 × 10 s

14 −1
V = 4.87 × 10 s

14 −1
Frequency of emission ( v) = 4.87 × 10 s
8 −1
(b) Velocity of radiation, (c) = 3.0 × 10 ms

Distance travelled by this radiation in 30s

8 −1
= (3.0 × 10 ms ) (30s)

9
= 9.0 × 10 m

(c) Energy of quantum (E) = hv

−34 14 −1
(6.626 × 10 Js) (4.87 × 10 s )

−20
Energy of quantum (E) = 32.27 × 10 J

−20
(d) Energy of one photon (quantum) = 32.27 × 10 J

−20
Therefore, 32.27 × 10 J of energy is present in 1 quantum.

Number of quanta in 2J of energy

2J
=
−30
32.27×10 J

18
= 6.19 × 10 J

18
= 6.2 × 10

Page : 71 , Block Name : Exercise

Q2.48 In astronomical observations, signals observed from the distant stars are generally weak. If
the photon detector receives a total of 3.15 × 10 J from the radiations of 600 nm, calculate the
−18

number of photons received by the detector.

Answer. From the expression of energy of one photon (E),
hc
E =
λ

Where,

λ = wavelength of radiation

h = Planck's constant

c = velock'ty of radiation

Substituting the values in the given expression of E:
−34 8 −1
(6.626×10 Js)(3×10 ms )

= −9
(600×10 m)

−19
E = 3.313 × 10 J

−19
Energy of one photon = 3.313 × 10 J

−18
Number of photons received with 3.15 × 10 J energy

−18
= 9.15 × 10 J

≈ 10

Page : 72 , Block Name : Exercise

Q2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation
source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns
and the number of photons emitted during the pulse source is 2.5 × 10 . calculate the energy of 15

the source.

Answer. Frequency of radiation (v),
1
v =
−9
2.0 × 10 s
8 −1
v = 5.0 × 10 s

Energy (E) of source Nhv
where, N = number of photons emitted
h = Planck's constant
v = frequency of radiation Substituting the values in the given expression of (E) :

E = (2.5 x 10 ) (6.626 x 10−34 JS) (5.0 x 10 )
15 8

E = 8.282 x 10 −10
J
Hence, the energy of the source (E) is 8.282 x 10 −10
J.

Page : 72 , Block Name : Exercise

Q2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm.
Calculate the frequency of each transition and energy difference between two excited states.

Answer. Missing

Page : 72 , Block Name : Exercise

Q2.51 The work function for caesium atom is 1.9 eV. Calculate
(a) the threshold wavelength and
(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength
500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer. It is given that the work function (Wo) for caesium atom is 1.9 eV.
hc
W0 =
λ0

(a) From the expression,

hc
λ0 =
W0

Where,

λ0 = threshold wavelength
h = Planck's constant
c = velocity of radiation
Substituting the values in the given expression of (λo):
−34 8 −1
(6.626 × 10 Js) (3.0 × 10 ms )
λ0 =
−19
1.9 × 1.602 × 10 J
−7
λ0 = 6.53 × 10 m

Hence, the threshold wavelength λ is 653 nm. 0

(b) From tie expression, W 0 = hv0 , we get:
W0
v0 =
h

Where,

V0 = threshold frequency

h = Planck's constant

Substituting the values in the given expression of v0.
−19
1.9×1.602×10 J
v0 = −34
6.626×10 Js

−19
(1eV = 1.602 × 10 J)

14 −1
V0 = 4.593 × 10 s

Hence, the threshold frequency of radiation (v ) is 4.593 x 10 0
14
s
−1

(c) According to the question:
Wavelength used in irradiation(λ) = 500 nm
Kinetic energy = h (v − v ) 0

1 1
= hc( − )
λ λ0

−14 4 λ0 −λ
−1
= (6.626 × 10 Js) (3.0 × 10 ms )( )
2λ0

−9
(653 − 500)10 m
−24
= (1.9878 × Jm) [ ]
−18 2
(653)(500)10 m

−26 9
(1.9878 × 10 ) (153 × 10 )
= J
(653)(500)

−20
= 9.3149 × 10 J

Kinetic energy of the ejected photoelectron = 9.3149 × 10 −20
J
1 2 −20
since K. E = mv = 9.3149 × 10 J
2
5 −1
v = 4.52 × 10 ms

5 −1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 10 ms .

Page : 72 , Block Name : Exercise

Q2.52 Following results are observed when sodium metal is irradiated with different wavelengths.
Calculate (a) threshold wavelength and, (b) Planck’s constant.

Answer.
−4
(a) Assuming the threshold wavelength to be λ0 nm (= λ0 × 10 m) , the kinetic energy

of the radiation is given as:

1 2
h (v − v0 ) = mv
2

Three different equalities can be formed by the given value as:

1 1 1 2
hc ( − ) = mv
λ λ0 2

1 1 1 −5 −2 −1
hc( 9
− ) = m (2.55 × 10 × 10 ms )
−9 2
500×10 λ0 ×10 m

hc 1 1 1 2
−3 −1
[ − ] = m(2.55 × 10 ms )
−9 500
10 m λ0 2

hc 2
1 1 1 +3 −1
[ − ] = m(3.45 × 10 ms )
−9 450
10 m λa 2

hc 2
1 1 1 +3 −1
[ − ] = m(5.35 × 10 ms )
−9 400
10 m λv 2

Dividing equation (3) by equation (1):
λ −400
0
2
[ ] +3 −1
400λ (5.35×10 ms )
0
= 2
λ −500 +3
0 −1
[ ] (2.55×10 ms )
500λ
0

2
5λ0 −2000 5.35 28.6225
= ( ) =
4λ0 −2000 2.55 6.5025

5λ0 −2000
= 4.40177
4λ11 −2000

17.6070λ0 − 5λ0 = 8803.537 − 2000

6805.537
λ0 =
12.607

λ0 = 539.8nm

λ0 = 540nm

∴ Threshold wavelength (λ0 ) = 540nm

Note: part (b) of the question is not done due to the incorrect values of velocity given in

the question.

2
5λ0 −2000 5.35 28.6225
= ( ) =
4λ0 −2000 2.55 6.5025

5λ0 −2000
= 4.40177
4

17.6070λ0 − 5λ0 = 8803.537 − 2000

6805.537
λ0 =
12.607

λ0 = 539, 8nm

λ0 = 540nm

Page : 72 , Block Name : Exercise

Q2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect
experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used.
Calculate the work function for silver metal.

Answer. From the principle of conservation of the mergv of an incident photon (E) is equal to the
sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,
E = W0 + K. E

⇒ W0 = E − K. E

Energy of incident photon (E) Where,
c = velocity of radiation
h = Planck's constant
λ= wavelength of radiation

Substituting the values in the given expression of E:
−34 8 −1
(6.626 × 10 Js) (3.0 × 10 ms )
E =
−9
256.7 × 10 m
−19
= 7.744 × 10 J
−19
7.744 × 10
= eV
−19
1.602 × 10

E = 4.83eV

The potential applied to silver metal changes to kinetic energy (K.E) Of the photoelectron. Hence,

KE = 0.35V

K. E = 0.35eV

∴ Work function, W0 = E − K .E

= 4.83eV − 0.35eV

= 4.48eV
2
5λ0 −2000 5.35 28.6225
= ( ) =
4λ0 −2000 2.55 6.5025

5λ0 −2000
= 4.40177
4λ0 −2000

17.6070λ0 − 5λ0 = 8803.537 − 2000

6805.537
λ0 =
12.607

λ0 = 539.8nm

λ0 = 540nm

Page : 72 , Block Name : Exercise

Q2.54 If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound electrons
is ejected out with a velocity of 1.5 × 10 ms , calculate the energy with which it is bound to the
7 −1

nucleus.

Answer.
Energy of incident photon (E) is given by,

hc
E =
λ

−34 8 −1
(6.626×10 Js)(3.0×10 ms )

= −12
(150×10 m)

−15
= 1.3252 × 10 J

−16
= 13.252 × 10 J

Energy of the electron ejected (K.E)

1 2
= mc v
2

1 2
−31 7 −1
= (9.10939 × 10 kg) (1.5 × 10 ms )
2

−17
= 10.2480 × 10 J

−16
= 1.025 × 10 J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E − K. E

−16 −16
= 13.252 × 10 J − 1.025 × 10 J

−16
= 12.227 × 10 J
−16
12.227 × 10
= eV
−19
1.602 × 10
3
=7.6 × 10 eV
2
5λ0 − 2000 5.35 28.6225
= ( ) =
4λ0 − 2000 2.55 6.5025

5λ0 − 2000
= 4.40177
4λη − 2000

17.6070λ0 − 5λ0 = 8803.537 − 2000

6805.537
λ0 =
12.607

λ0 = 539.8nm

λ0 = 540nm

Page : 72 , Block Name : Exercise

Q2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can
be represented as v = 3.29 × 10 (Hz) [1/3 − 1/n ]. Calculate the value of n if the transition is
15 2 2

observed at 1285 nm. Find the region of the spectrum.

Wavelength of transition = 1285nm

−9
= 1285 × 10 m( Given )
Answer.
15 1 1
v = 3.29 × 10 ( 2
− 2
)
3 n
( Given )

c
v =
λ

since
8 −1
3.0×10 ms
=
−9
1285×10 m

−9
v = 2.33 × 10 m

Substituting the value of v in the given expression,

15 1 1 14
3.29 × 10 ( − 2
) = 2.33 × 10
9 n

14
1 1 2.33×10
− =
9 2 15
n 3.29×10

1 −1 1
− 0.7082 × 10 =
9 2
n

1 −1 −1
⇒ = 1.1 × 10 − 0.7082 × 10
2
n

1 −2
= 4.029 × 10
2
n

1
n = √ −2
4.029×10

n = 4.98

n ≈ 5

Hence, for the transition to be observed at 1285nm, n = 5

The spectrum lies in the infrared region.

Page : 72 , Block Name : Exercise

Q2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius
1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region
of the spectrum.

Answer. The radius of the n th
orbit of hydrogen-like particles is given by,
2
0.529n ˙
r = A
Z

2
52.9n
r = pm
Z

For radius (r1 ) = 1.3225nm

−9
= 1.32225 × 10 m

−12
= 1322.25 × 10 m

= 1322.25pm

Page : 72 , Block Name : Exercise

Q2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope
often used for the highly magni ed images of biological molecules and other type of material. If
the velocity of the electron in this microscope is 1.6 × 10 ms 1, calculate de Broglie wavelength
6 −1

associated with this electron.

Answer. From de Broglie’s equation,
h
λ =
mv
−34
6.626 × 10 Js
λ =
−31 6 −1
(9.10939 × 10 kg) (1.6 × 10 ms )

−10
=4.55 × 10 m

λ = 455pm

∴ de Broglie's wavelength associated with the electron is 455pm

Page : 72 , Block Name : Exercise

Q2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the
determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the
characteristic velocity associated with the neutron.

From de Broglie's equation,

h
λ =
mv
Answer. h
v =
mλ

Where,

v = velocity of particle (neutron)

h = Planck's constant

m = mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of velocity (v) ,
−34
6.626×10 Js
v =
−27 −12
(1.67493×10 kg)(800×10 m)

1
r1 Z
2
n =
1 52.9

2 1322.25Z
n =
1 52.9

Similarly,

2 211.6Z
n =
2 52.9
2
n
1 1322.5
=
2 211.6
n
2

2
n
1
= 6.25
2
n
2

n1
= 2.5
n2

n1 25 5
= =
n2 10 2

⇒ r1 = 5 and n2 = 2

th nd
Thus, the transition is from the 5 orbit to the 2 orbit. It belongs to the Balmer series.

¯
¯¯
Wave number (v ) for the transition is given by,

7 −1 1 1
1.097 × 10 m ( − 2
)
2
2 5

7 −1 2l
= 1.097 × 10 m ( )
100

6 −1
= 2.303 × 10 m

∴ Wavelength (λ) associated with the emission transition is given by,

^ 1
λ =
¯
¯¯
v

1
= 6
2.303×10 m−1

−6
= 0.434 × 10 m

λ = 434nm

Page : 72 , Block Name : Exercise

Q2.59 If the velocity of the electron in Bohr’s rst orbit is 2.19 × 10 ms 6 −1
. calculate the de Broglie
wavelength associated with it.

Answer. According to de Broglie's equation,

Where,
λ =wavelength associated with the electron

h = Planck's constant

m = mass of electron
V = velocity of electron
Substituting the values in the expression of λ :
−34
6.626 × 10 Js
=
−31 6 −1
(9.10939 × 10 kg) (2.19 × 10 ms )

100
−10 −10
= 3.32 × 10 m = 3.32 × 10 m ×
100
−12
λ = 332 × 10 m

∴ Wavelength associated with the electron = 332pm

Page : 72 , Block Name : Exercise

Q2.60 The velocity associated with a proton moving in a potential difference of 1000 V is
. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the
5 −1
4.37 × 10 ms

wavelength associated with this velocity.

Answer.
According to de Broglie's expression,

h
λ =
mv

Substituting the values in the expression,
−34
6.626×10 Js
λ =
5
(0.1kg)(4.37×10 ms−1 )

−38
λ = 1.516 × 10 m

Page : 72 , Block Name : Exercise

Q2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the
uncertainty in the momentum of the electron. Suppose the momentum of the electron is
h/4π m × 0.05nm, is there any problem in de ning this value.

Answer. From Heisenberg’s uncertainty principle,
h 1 h
Δx × Δp = ⇒ Δp = ⋅
4π Δx 4π

Where,
Δx = uncertainty in position of the electron

Δp = uncertainty in momentum of the electron

Substituting the values in the expression of Δp:
−44
1 6.626 × 10 Js
Δp = ×
0.002nm 4 × (3.14)

−34
1 6.626 × 10 Js
= ×
−12
2 × 10 m 4 × 3.14

−23 −1
=2.637 × 10 Jsm

−23 −1 2 −1
Δp = 2.637 × 10 kgms (1J = 1kgms s )

−23 −1
∴ Uncertainty in the momentum of the electron = 2.637 × 10 kgms

h
=
4πm ×0.05nm
−34
6.626×10 Js
= −11
4×3.14×5.0×10 m

−11
= 1 × 3.14 × 5.0 × 10 m

Since the magnitude of the actual momentum is smaller than the uncertainty, the value

cannot be defined.

Page : 72 , Block Name : Exercise

Q2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing
energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, m = −2, m = −1/2 \)
l s

2. n = 3, l = 2, m = 1, m = +1/2
l s

3. n = 4, l = 1, m = 0, m = +1/2
l s

4. n = 3, l = 2, m = −2, m = −1/2
l s

5. n = 3, l = 1, m = −1, m = +1/2
l s

6. n = 4, l = 1, m = 0, m = +1/2
l s

Answer. For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2 , the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals
respectively.
Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1(4d).

Page : 72 , Block Name : Exercise

Q2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in
3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective
nuclear charge ?

Answer. Nuclear charge experienced by an electron (present in a multi-electron atom) is
dependant upon the distance between the nucleus and the orbital, in which the electron is present
As the distance increases, the effective nuclear charge also decreases. Among p-orbitals, 4p
orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in
the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by
electrons present in the 20 and 30 orbitals along with the s-orbitals. Therefore, they will
experience the longest nuclear charge.

Page : 73 , Block Name : Exercise

Q2.64 Among the following pairs of orbitals which orbital will experience the larger effective
nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

Answer. Nuclear charge is de ned as the net positive charge experienced by an electron in the
orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge
experienced by the electron (s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to
the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

Page : 73 , Block Name : Exercise

Q2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will
experience more effective nuclear charge from the nucleus ?

Answer. Nuclear charge is de ned as the net positive charge experienced by an electron in a multi-
electron atom.
The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while
aluminum has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium,
Which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience
more effective nuclear charge than aluminium.

Page : 73 , Block Name : Exercise

Q2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Answer. (a) Phosphorus (P):
Atomic number = 15
The electronic con guration of p is: 1s 2s 20 3s 3p
2 2 6 2 3

The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.
(b) Silicon (Si):
Atomic number = 14
The electronic con guration of Si is:
The orbital Picture of Si can be represented as:

From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr):
Atomic number = 24
The electronic con guration of Cr is:
2 2 6 2 6 1 5
1s 2s 2p 3s 3p 4s 3d

The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe):
Atomic number = 26
The electronic con guration is:
2 2 6 2 6 2 6
1s 2s 2p 3s 3p 4s 3d

The orbital picture of Iron is:

From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr):
Atomic number = 36
The electronic con guration is:
2 2 6 2 6 2 10 6
1s 2s 2p 3s 3p 4s 3c 4p

The orbital picture of krypton is:

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Page : 73 , Block Name : Exercise

Q2.67 (a) How many subshells are associated with n = 4 ?
(b) How many electrons will be present in the subshells having ms value of –1/2 for n = 4 ?

Answer. (a) n = 4 (Given)
For a given value of 'n', I can have values from zero to (n - 1).
Thus, four subshells associated with n = 4, which are s, p, d and f.
(b) Number of orbitals in the n shell = n
th 2

For n = 4
Number Of orbitals = 16
If each orbital is taken fully, then it Will have 1 electron With m, value of −
1

2

∴ Number of electrons with m, value of (− 1
) .
2
is 16

Page : 73 , Block Name : Exercise