The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy is

Options

A. 3.4 eV
B. 6.8 eV
C. 10.2 eV
D. zero

Answer

C

Explanation

Given ground state energy of hydrogen atom
`E_(1)=-13.6eV`
Energy of electron in first excited state (ie, n=2)
`E_(2)=-(13.6)/((2)^(2))eV`
Therefore, excitation energy
`DeltaE=E_(2)-E_(1)`
`=-(13.6)/(4)-(13.6)`
`=-3.4+13.6=10.2eV`