Question

Calculate the frequency of the light emitted by ahydrogen atom

during a transition of its electron from the n = 4 tothe n = 1
principal energy level.

Answer 1

Answer:
`0.31* 10^(16)Hz`.

Step-by-step explanation:

`E=(hc)/(\lambda)`

`\lambda` = Wavelength of radiation

E= energy

Using Rydberg's Equation:

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )* 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

`(1)/(\lambda)=10973731.6m^(-1)\left((1)/(1^2)-(1)/(4^2) \right )* 1`

`\lambda=9.7* 10^(-8)m`

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

`\\u=(c)/(\lambda)`

where,

`\\u` = frequency of the wave = ?

c = speed of light =
`3* 10^8ms^(-1)`

`\lambda` = wavelength of the wave =
`9.7* 10^(-8)m`

`\\u=(3* 10^8ms^(-1))/(9.7* 10^(-8)m)`

`\\u=0.31* 10^(16)s^(-1)=0.31* 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* 10^(16)Hz`.