If, an electron in hydrogen atom jumps from an orbit of lelvel n=3 to an orbit of level n=2, emitted radiation has a freqwuency (R= Rydbertg's contant ,c = velocity of light)

Options

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

Answer

D

Explanation

`1/lamda = R[1/2^2 - 1/3^2] = R[1/4 - 1/9] = (5R)/(36)`
`therefore "Frequency " v = C / lamda = (C.5R)/(36) = (5RC)/(36)`