Question

What will be the wavelength of a photon that is emitted when an electron of a hydrogen atom undergoes a transition from n=7 to n=4?

Answer 1

Answer: $2.16\times 10^{-6}m$

Explanation:

Using Rydberg's Equation:

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

Where,

= Wavelength of radiation

$R_H$ = Rydberg's Constant  = $10973731.6m^{-1}$

$n_f$ = Higher energy level = 7  

= Lower energy level = 4

Z= atomic number = 1 (for hydrogen)

$\frac{1}{\lambda}=10973731.6m^{-1}\left(\frac{1}{4^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=461904.5$

$\frac{1}{\lambda}=461904.5$

$\lambda=2.16\times 10^{-6}m$

Thus wavelength is $2.16\times 10^{-6}m$