Objective Questions of Atomic Structure


For a p-electron, orbital angular moment is

(A) √2h                               (B)   h

(C) √6h                              (D)   2h


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

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

∴  (A)


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

(A)    H                              (B)    He+

(C)    Li2+                           (D)    Na+


Bohr theory is not applicable to multi electron species

∴ (D)


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


r = n2h2/ 4π2mZe

∴ r2/r3 = 22/32

∴ r3  = 9/4r2

∴  (C)


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

(A)    2                              (B)    3

(C)    4                              (D)    1


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)

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

(A)    16                             (B)    4

(C)    2                              (D)    1


En =  - RH/n2

∴ - RH/n2 = - RH/16 

i.e. for 4th sub-shell 

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

∴  Degeneracy is 16


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


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)


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


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)


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


They are distinguished by their spin.

∴ (D) 


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-


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)


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


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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