Objective Questions of Atomic Structure

Problem:

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

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

Bohr theory is not applicable to multi electron species

If the radius of 2^nd 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/ 4π2mZe² 

∴ r₂/r₃ = 2²/3²

∴ r₃  = 9/4r₂

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

Circumference of 3^rd orbit = 2πr₃

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

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

By de-Broglie equation, 

∴ λ = h/mv

∴ λ   = 2πr₃/3

∴ 2λr₃  = 3λ

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

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

E_n =  - R_H/n²

∴ - R_H/n² = - R_H/16 

i.e. for 4^th sub-shell 
      

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

(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?

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

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

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

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