Question

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =3 to =1.

Answer 1

Answer:

$1.936\times 10^{-18}\ \text{J}$

Explanation:

$R_h$ = Rydberg constant = $2.178\times 10^{-18}\ \text{J}$

$n_i$ = Initial shell = 3

$n_f$ = Final shell = 1

We have the relation

$\Delta E=R_h(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\\\Rightarrow \Delta E=2.178\times 10^{-18}(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\\Rightarrow \Delta E=1.936\times 10^{-18}\ \text{J}$

The energy of the photon emitted here is $1.936\times 10^{-18}\ \text{J}$.