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The total energy of an electron in the first excited state of hydrogen atom is about $-3.4$ eV. Its kinetic energy in this state is -
A
$-6.8$ eV
B
3.4 eV
C
6.8 eV
D
$-3.4$ eV
Detailed Solution
In a Bohr orbit the kinetic energy is $K = \dfrac{1}{2}\dfrac{ke^2}{r}$ and the potential energy is $U = -\dfrac{ke^2}{r}$.
So the total energy is $E = K + U = -\dfrac{1}{2}\dfrac{ke^2}{r} = -K$.
Thus K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
K.E. $= -(-3.4) = +3.4$ eV
(The potential energy in this state is $-6.8$ eV.)
So the total energy is $E = K + U = -\dfrac{1}{2}\dfrac{ke^2}{r} = -K$.
Thus K.E. $= -$(total energy) and P.E. $= 2 \times$ (total energy).
K.E. $= -(-3.4) = +3.4$ eV
(The potential energy in this state is $-6.8$ eV.)