Question

When an excited electron falls from n = 3 to n = 2, a red radiation in the Balmer series emitted. What is the frequency of this photon?

A. 6.63 x 10-34 Hz
B. 2.19 x 1016 Hz
C. 5.28 x 1011 Hz
D. 4.57 x 1014 Hz

Answer 1

The frequency of this photon is
`4.57*10^(14)\ Hz`

(D) is correct option.

Step-by-step explanation:

Given that,

Excited states,

`n=3`

`n=2`

We need to calculate the wavelength

Using formula for energy

`E=13.6((1)/(n^2)-(1)/(m^2))`

`E=13.6((1)/(4)-(1)/(9))`

`E=1.888\ eV`

`E=1.888*1.6*10^(-19)\ J`

`E=3.0208*10^(-19)\ J`

We need to calculate the frequency

Using formula of frequency

`E=hf`

`f=(E)/(h)`

Where, E =energy

`f=(3.0208*10^(-19))/(6.63*10^(-34))`

`f=4.57*10^(14)\ Hz`

Hence, The frequency of this photon is
`4.57*10^(14)\ Hz`