Monochromatic radiation of wavelength `lambda` is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiations of six different wavelength . Find the wavelength `lambda`.

Explanation

If electrons are excited up to the state n then number of possible transitions or the wavelengths would be n(n - 1)/2. Here 6 different wavelengths are emitted hence
n(n-1)/2= 6
On solving we get n = 4. Hence electrons are shifted up to the state n = 4.
Energy in the `4^(th)` state can be written as follows:
`E_4=E_1/4^2=-13.6/16`=-0.85 eV
Energy needed to make the transit from ground state to n = 4 state.
`DeltaE=E_4-E_1=(-0.85)-(-13.6)=12.75` eV
If `lambda` is wavelength of emitted radiation then
`hc//lambda=12.75xx1.6xx10^(-19)`
`rArr lambda=(hc)/(12.75xx1.6xx10^(-19))=(6.63xx10^(-34)xx3xx10^8)/(12.75xx1.6xx10^(-19))`
`=0.975xx10^(-7)` m = 97.5 nm