Solved Questions of Atomic Structure Part I

Problem 1:

Why Bohr’s orbits are called stationary states?

This is because the energies of orbits in which the electrons revolve are fixed.

Problem 2:

Explain why the electronic configuration of Cu is 3d¹⁰4s¹ and not 3d⁹4s².

In the 3d¹⁰4s¹ the d-sub shell is completely filled which is more stable.

Problem 3:

Fe³⁺ ion is more stable than Fe²⁺ ion. Why?

In Fe³⁺ ion 3d sub shell is half filled hence more stable configuration.

Problem 4:

Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.005 nm.

Problem 5:

Give one example of isodiapheres.

Isodiapheres have same difference between the number of neutrons and protons. For example

Problem 6:

Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He⁺. [Neglect reduced mass effect].

= 4R [ 36 - 16/36 x 16 ] = 5R/36

v^-_H = R x 1² [ 1/2² - 1/n²]

∴ v^-_He⁺ = v^-_H

 5R/36 = R/4 - R/n²

On solving above equation

n² = 9               

Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as that of 6 → 4 transition in He⁺. 

Problem 7:

Calculate ionization potential in volts of (a) He⁺ and (b) Li²⁺       

I.E. = 13.6Z²/n² 

=  13.6 x Z² \1² [Z =2 for He⁺]

Similarly for Li²⁺ = 13.6 x 3²/1²

Problem 8:

Calculate the ratio of K.E and P.E of an electron in an orbit?

K.E. = Ze²/2r

P.E. = -Ze²/r

∴ K.E/P.E = - 1/2 

Problem 9:

How many spectral lines are emitted by atomic hydrogen excited to 
n^th energy level?

Thus the number of lines emitted from nth energy level

∑n =  n(n+1)/2      

∴ ∑ (n – 1) = ( n-1) (n-1+1)/2  = (n-1) (n)/2 

Number of spectral lines that appear in hydrogen spectrum when an electron jumps from n^th energy level = n (n-1)/2