Question

A hydrogen atom in an excited state absorbs a photon of wavelength 411 nm. What were the initial and final states of the hydrogen atom?

Answer 1

Answer:

The initial and final states of the hydrogen atom were n=2 and n=6 respectively.

Explanation:

We must first obtain the energy of the photon;

E= hc/λ

where;

h= Plank's constant = 6.6 * 10^-34 JS

c= speed of light = 3* 10^8 m/s

λ = wavelength of light= 411 nm = 411* 10^-9 m

Substituting values;

E = 6.6 * 10^-34 * 3* 10^8 / 411* 10^-9

E = 4.8 * 10^-19 J or 3.0 eV

But ;

En = 13.6/n^2

So E = En final - En initial

3.0  = -13.6(1/n^2final - 1/n^2initial)

If we substitute n^2final = 6 and n^2 initial = 2 then the RHS becomes approximately equal to the LHS

Therefore the initial and final states of the hydrogen atom were n=2 and n=6 respectively.