Question

A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 6 excited state. The ion returns to the ground state by emitting FIVE photons ONLY. What is the wavelength of the second highest energy photon?

Answer 1

Given singly ionized helium atom in the ground state jumps to n= 6 exited state

emitting the FIVE photons , the possible

Tansitions are

n=6 ---> n= 5

n=5 ---> n= 4

n=4 ---> n= 3

n=3 ---> n= 2

n=2 ---> n= 1

, among these the second highest energy state is n = 3 ---> n=2 whose wavelength is givne by
  
  

   1/lambda = R*z^2(1/n2^2-1/n1^2)

   1/lambda = 1.097*10^7*2^2(1/2^2-1/3^2)

   lambda = 1.6408*10^-7 m

   lambda = 164.08 nm