Question

A hydrogen atom in an excited state emits a photon of frequency ν = 3.084 x 1015 s-1. If the electron returns to the ground state, in which level was it before the photon was emitted?

Answer 1

Answer: The photon was emitted from the 4th energy level

Explanation:

To calculate the wavelength of light, we use the equation:

$\lambda=\frac{c}{\nu}$

where,

$\lambda$ = wavelength of the light  

c = speed of light = $3\times 10^8m/s$

$\nu$ = frequency of light = $3.084\times 10^{15}s^{-1}$

Putting the values in above equation, we get:

$\lambda=\frac{3\times 10^8m/s}{3.084\times 10^{15}s^{-1}}=9.73\times 10^{-8}m$

To calculate the initial level, we use the equation given by Rydberg's equation, which is:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

where,

$\lambda$ = Wavelength of radiation  $9.73\times 10^{-8}m$

$R_H$ = Rydberg's Constant  = $1.097\times 10^7m^{-1}$

$n_f$ = Final energy level =  1  

$n_i$= Initial energy level =  ?

Putting the values, in above equation, we get

$\frac{1}{9.73\times 10^{-8}m}=1.097\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{n_i^2} \right )$

$n_i=4$

Hence, the photon was emitted from the 4th energy level