Calculate the wavelength of radiation emitted when `He^(+)` makes a transition from the state `n = 3` to the state `n = 2`

Options

A. `16400nm`
B. `164.0nm`
C. `1.64nm`
D. `0.164nm`

Answer

B

Explanation

The wavelength `lambda` is given by
`(1)/(lambda) = RZ^(2)((1)/(n^(2)) - (1)/(m^(2))`
`= 4R((1)/(4) - (1)/(9)) = (5)/(9)R`
`or lambda = (9)/(5R) = (9)/(5 xx 1.09737 xx 10^(2)m^(-1)) = 164.0nm`