Question

an electron jumps in lyman series from 3rd shell of H-atom.find the energy released.

Answer 1

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].

 

Solution:-

v-He+ = RZ2 [ 1/42 - 1/62]

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

v-H = R x 12 [ 1/22 - 1/n2]

So, v-He+ = v-H

5R/36 = R/4 - R/n2

On solving above equation

n2 = 9               

Thus, n = 3

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