For the hydrogen atom, the energy of radiation emitted in the transition from 4th excited state to 2nd excited state, according to Bohr's theory is

Options

A. 0.567 eV
B. 0.667 eV
C. 0.967 eV
D. 1.267 eVs

Answer

B

Explanation

Using `E_(n)=(-13.6)/(n^(2))eV`
According to given question, ltbgt `E_(4th)=-(13.6)/((4)^(2))=-(13.6)/(16)eV=-0.85eV`
And `E_(2nd)=(13.6)/((2)^(2))=(-13.6)/(4)eV=-3.4eV`
`E_(3rd)=(-13.6)/((3)^(2))=-(13.6)/(9)=-1.51eV`
Energy of radiation emitted in the transition
`DeltaE=E_(4th)-E_(2nd)`
`=-0.85+3.4=2.55eV`
`DeltaE=E_(3rd)-E_(2nd)`
`=-15.1+3.4=1.89eV`
`DeltaE=E_(4th)-E_(3rd)`
`=-0.85+1.51=0.66eV`