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The total energy of electron in the ground state of hydrogen atom is $-13.6$ eV. The kinetic energy of an electron in the first excited state is
A
6.8 eV
B
13.6 eV
C
1.7 eV
D
3.4 eV
Detailed Solution
Total energy in the $n$th orbit: $E_n = \dfrac{-13.6}{n^2}$ eV
The first excited state corresponds to $n = 2$.
Energy in the first excited state $= \dfrac{-13.6}{2^2} = -3.4$ eV
In a Bohr orbit, K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
$\therefore$ K.E. $= -(-3.4) = +3.4$ eV
The first excited state corresponds to $n = 2$.
Energy in the first excited state $= \dfrac{-13.6}{2^2} = -3.4$ eV
In a Bohr orbit, K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
$\therefore$ K.E. $= -(-3.4) = +3.4$ eV