Question

Calculate the frequency of the light emitted by ahydrogen atom

Answer 1

Answer:  $0.31\times 10^{16}Hz$.

Explanation:

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

$\lambda$ = Wavelength of radiation

E= energy

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,

$\lambda$ = Wavelength of radiation  = ?

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 4

$n_i$= Lower energy level = 1

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

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

$\lambda=9.7\times 10^{-8}m$

The relationship between wavelength and frequency of the wave follows the equation:

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

where,

$\nu$ = frequency of the wave  = ?

c = speed of light  = $3\times 10^8ms^{-1}$

$\lambda$ = wavelength of the wave = $9.7\times 10^{-8}m$

$\nu=\frac{3\times 10^8ms^{-1}}{9.7\times 10^{-8}m}$

$\nu=0.31\times 10^{16}s^{-1}=0.31\times 10^{16}Hz$

The frequency of the light emitted by a hydrogen atom  during a transition of its electron from the n = 4 to the n = 1  principal energy level is $0.31\times 10^{16}Hz$.