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In hydrogen atom, energy of first excited state is – 3.4 eV. Then find out KE of same orbit of hydrogen atom : -
A
+3.4 eV
B
+6.8 eV
C
–13.6 eV
D
+13.6 eV
Detailed Solution
In the Bohr model, for an electron in an orbit: potential energy $U = -\dfrac{kZe^2}{r}$, kinetic energy $KE = \dfrac{kZe^2}{2r}$.
Total energy $E = KE + U = -\dfrac{kZe^2}{2r} = -KE$
So $KE = -E$ (and $U = 2E$).
For the first excited state, $E = -3.4$ eV, so $KE = +3.4$ eV.
(Its potential energy is $-6.8$ eV.)
Total energy $E = KE + U = -\dfrac{kZe^2}{2r} = -KE$
So $KE = -E$ (and $U = 2E$).
For the first excited state, $E = -3.4$ eV, so $KE = +3.4$ eV.
(Its potential energy is $-6.8$ eV.)