Question

A photon is emitted when a hydrogen atom undergoes a transition from the n = 8 state to the n = 4 state. Calculate values for the following. (a) the wavelength nm (b) the frequency Hz (c) the energy of the emitted photon eV

Answer 1

a) λ=2 μm

b)
`f= 1.5 * 10^(14)`

c)E=0.61 eV

Step-by-step explanation:

Given that

Hydrogen atom undergoes from n=8 to n= 4 state.

a)To find wavelength

We know that

`(1)/(\lambda)=R_H\left((1)/(n_2^2)-(1)/(n_1^2)\right)`

Where

`R_H=1.09* 10^7\ m^(-1)`

Now by putting the values

`(1)/(\lambda)=R_H\left((1)/(n_2^2)-(1)/(n_1^2)\right)`

`(1)/(\lambda)=1.09* 10^7\left((1)/(4^2)-(1)/(8^2)\right)`

`(1)/(\lambda)=510937.5`

λ=2 μm

b)For frequency

C= f x λ

`f=(3* 10^8)/(2* 10^(-6))\ Hz`

`f= 1.5 * 10^(14)`

c)To find energy

E= hf

`E=6.6* 10^(-34)*1.5 * 10^(14)\ J`

`E=9.9* 10^(-20)\ J`

`E=(9.9* 10^(-20))/(1.6* 10^(-19))\ eV`

E=0.61 eV