Which of the following transitions of an electron in hydrogen atom emits of the lowest wavelength?

Options

A. `n_(2)=oo`to `n_(2)`
B. `n_(2)=4`to`n_(1)=3`
C. `n_(2)=2`to`n_(1)`
D. `n_(2)=5`to`n_(1)=3`

Answer

c

Explanation

`n alpha1/(DeltaE)`
If `n` value increases,`Delta E(E_n-E_(n-1))` i.e,`E_2-E_1gtE_3-E_2gt......`
`DeltaE=(hc)/lambda impliesDeltaEalpha1/lambda`
Hence ,`n_2=2` to `n_1=1` has lowest wavelength.