Question

calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 3 to n = 1 energy level based on the bohr theory

Answer 1

`\\u=2.92* 10^(15)\ Hz`

Step-by-step explanation:

The expression for the energy of an electron in the nth orbit is:-

`E_n=-2.18* 10^(-18)* (1)/(n^2)\ Joules`

For transitions:

`Energy\ Difference,\ \Delta E= E_f-E_i =-2.18* 10^(-18)((1)/(n_f^2)-(1)/(n_i^2))\ J=2.18* 10^(-18)((1)/(n_i^2) - (1)/(n_f^2))\ J`

`\Delta E=2.18* 10^(-18)((1)/(n_i^2) - (1)/(n_f^2))\ J`

Given,
`n_i=3\ and\ n_f=1`

`\Delta E=2.18* 10^(-18)((1)/(3^2) - (1)/(1^2))\ J`

`\Delta E=-1.94* 10^(-18)\ J` (negative sign indicates energy release)

Also,
`E=h* \\u`

Where,

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

`1.94* 10^(-18)=6.626* 10^(-34)* \\u`

`\\u=2.92* 10^(15)\ Hz`