If an electron in hydrogen atom jumbs from an orbit of level n = 3 to an orbit of level n = 2, the emitted radiation has a frequency

Options

A. `(RC)/(25)`
B. `(5RC)/36`
C. `(3RC)/27`
D. `(8RC)/9`

Answer

B

Explanation

When an electron jumps from higher level `n_(1)` to lower energy level `n_(2)` , the frequency of the emitted radiation is
` v = RC [1/(n_(2)^(2)) - 1/(n_(1)^(2))] `
` :. ` For n = 3 to n = 2 ,
` v = RC [ 1/(2^(2)) - 1/(3^(2))] = RC [1/4 -1/9]`
` = RC [ (9-4)/36] = (5RC)/36 `