Question

What is the energy (in J) of a photon that is ejected from a hydrogen atom when an electron relaxes from the 6th to the 3rd energy level ?

Answer 1

Answer: $1.8\times 10^{-19}J$

Explanation:

Using Rydberg's Equation for hydrogen atom:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level

$n_i$= Lower energy level

We have:

$n_f=6, n_i=3$

$R_H=1.09\times 10^7 m^{-1}$

$\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \left(\frac{1}{3^2}-\frac{1}{6^2} \right )$

$\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \frac{3}{36}$

$\frac{1}{\lambda}=0.0908\times 10^{7}m^{-1}$

$\lambda=11.0\times 10^{-7} m$

The relation between energy and wavelength of light is given by Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

E = energy of the light

h = Planck's constant

c = speed of light

$\lambda$ = wavelength of light

$E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{11.0\times 10^{-7}m}$

$E=1.8\times 10^{-19}J$

Thus energy of a photon that is ejected from a hydrogen atom when an electron relaxes from the 6th to the 3rd energy level is $1.8\times 10^{-19}J}$