Question

The electron in a hydrogen atom is an excited state makes a downward transition and a photon whose energy is 2.856eV is emited in the process. Between what two shells (giv principal quantum numbers) did the transition take place?

Answer 1

To resolve this problem we will use the expression of the hydrogen atom of Borh

E = - 13.606/n²

We write this expression for two different principal quantum numbers

E2 = -13.606 1/n₂²

E1 = -13.606 1/n₁²

E2 – E1 = - 13.606 ( 1/n₂^{2 –}1/n₁²)

2.856 = -13.606 ( 1/n₂^{2 –}1/n₁²)

- 2.856/13.606 = ( 1/n₂^{2 –} 1/n₁² )

- 0.2099 = ( 1/n₂^{2 –} 1/n₁² )

we must try various combinations

n₂ n₁   1/n₂²   1/n₁² (1/n₂^{2 –}1/n₁²)

2 1 0.25 1 -0.75

3 1 0.1111 1 -0.8888

4 1 0.0625 1 -0.9375

3 2 0.11 0.25 - 0.1389

4 2 0.0625 0.25 -0.1875

5 2 0.04 0.25 -0.21

4 3 0.0625 0.111 -0,0486

In examining the table we see that the quantum numbers n 2 = 5 and n1 = 2 are those that are closer to the desired transition

Results

n2= 5

n1 =2