Objective Questions of Atomic Structure

Problem:

For a p-electron, orbital angular moment is

(A) √2h                               (B)   h

(C) √6h                              (D)   2h

Solution:

Orbital angular momentum L = √l ( l + 1 )h where h = h/2π

∴ L for p electron = √1 ( 1 + 1 )h = √2h

∴  (A)

Problem:

For which of the following species, Bohr theory doesn’t apply

(A)    H                              (B)    He+

(C)    Li2+                           (D)    Na+

Solution:

Bohr theory is not applicable to multi electron species

∴ (D)
 

Problem:

If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be

(A)   4/9r2                             (B)    4r2

(C)   9/4r2                             (D)    9r2

Solution:

r = n2h2/ 4π2mZe

∴ r2/r3 = 22/32

∴ r3  = 9/4r2

∴  (C)

Problem:

Number of waves made by an electron in one complete revolution in 3rd Bohr orbit is

(A)    2                              (B)    3

(C)    4                              (D)    1

Solution:

Circumference of 3rd orbit = 2πr3

According to Bohr’s angular momentum of electron in 3rd orbit is

mvr3 = 3h/2π                or    h/mv = 2πr3/3

By de-Broglie equation, 

∴ λ = h/mv

∴ λ   = 2πr3/3

∴ 2λr = 3λ

i.e. circumference of 3rd orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in 3rd orbit is three.

∴ (B)
Problem:

The degeneracy of the level of hydrogen atom that has energy - RH/16  is

(A)    16                             (B)    4

(C)    2                              (D)    1

Solution:

En =  - RH/n2

∴ - RH/n2 = - RH/16 

i.e. for 4th sub-shell 
 1830_Solution.JPG     

i.e. 1 + 3 + 5 + 7 = 16

∴  Degeneracy is 16

Problem:

An electron is moving with a kinetic energy of 4.55 x 10–25 J. What will be de Broglie wavelength for this electron?

(A)    5.28 x 10–7 m               (B)    7.28 x 10–7 m

(C)    2 x 10–10 m                  (D)    3 x 10–5 m

 
Solution:

KE =  mv2 = 4.55 x 10–25

v2 =  2 x 4.55 x 10-25 / 9.1 x 10-31 = 1 x 106

v = 103 m/s

de Broglie wavelength λ = h/mv  = 6.626 x 10-34/ 9.1 x 10-31x 103= 7.28 x 10–7 m

∴ (B)

Problem:

Suppose 10–17J of energy is needed by the interior of human eye to see an object. How many photons of green light (λ = 550 nm) are needed to generate this minimum amount of energy?

(A)    14                             (B)    28

(C)    39                             (D)    42

Solution:

Let the number of photons required = n

n hc/λ = 10-16

n = 10-17 x λ /hc  = 10-17 x 550 x 10-9 / 6.626 x 10-34 x 3 x 108= 27.6 = 28 photons

∴ (B)

Problem:

The two electrons present in an orbital are distinguished by

(A)    Principal quantum number    (B)    azimuthal quantum number

(C)    Magnetic quantum number   (D)    spin quantum number

Solution:

They are distinguished by their spin.

∴ (D) 

Problem:

The velocity of electron in the ground state hydrogen atom is 
2.18 x 106 ms–1. Its velocity in the second orbit would be

(A) 1.09 x 106 ms–1              (B)    4.38 x 106 ms–1

(C) 5.5 x 105 ms–1                (D)    8.76 x 106 ms–1-

Solution:

We know that velocity of electron in nth Bohr’s orbit is given by

v  = 2.18 x 106 z/n  m/s

For H, Z = 1

... v1 = 2.18 x 106/1  m/s         

... v2 =  2.18 x 106/2  m/s = 1.09 x 106 m/s

∴ (A)

Problem:

The ionization energy of the ground state hydrogen atom is 
2.18 x 10–18J. The energy of an electron in its second orbit would be

(A)    –1.09 x 10–18 J             (B)    –2.18 x 10–18J

(C)    –4.36 x 10–18J              (D)    –5.45 x 10–19J

Solution:

Energy of electron in first Bohr’s orbit of H–atom

E = -2.18 x 10-18/n2  (Q ionization energy of H = 2.18 ´10–18J)

E2 = -2.18 x 10-18/22  J = –5.45 x 10–19J

∴ (D)
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