Calculate the wavelength of radiation emitted when an electron in a hydrogen atom makes a transition from an energy level with `n = 3` to a level with `n= 2`.

Options

A. 654.3 nm
B. 827.5 nm
C. 487.5 nm
D. 711.5 nm

Answer

A, C

Explanation

Here the possible transitions are from n=3 level to n=2 level, n=4 to n=2 level, n=5 to n=2 level and so on.
We know that energy for `n^(th)` level in hydrogen atom is given by the following relation :
`E_n=E_1/n^2`, where `E_1`=-13.6 eV
`E_2=(-13.6)/2^2=(-13.6)/4=-3.4` eV
`E_3=(-13.6)/3^2=(-13.6)/9`=-1.5 eV
`E_4=(-13.6)/4^2 =(-13.6)/16`=-0.85 eV
Highest wavelength `lambda_1` can be calculated as follows :
`lambda_1=(hc)/(E_3-E_2)=(6.63xx10^(-34)xx3xx10^8)/((-1.5+3.4)xx1.6xx10^(-19))`
`rArr 6.54xx10^(-7)` n = 654 nm
Second highest wavelength `lambda_2` can be calculated as follows :
`lambda_2=(hc)/(E_4-E_2)=(6.63xx10^(-34)xx3xx10^8)/((-0.85+3.4)xx1.6xx10^(-19))`
`=4.87xx10^(-7)`m = 487 nm