Compare the potential energy and kinetic energy of electrons in the 1st orbits of hydrogen and $H e^{+}$ ion.

Open in App

Solution

PE in H= $- 2 \times 13.6 \times \frac{z^{2}}{n^{2}} = - 2 \times 13.6 = - 27.2 e V P E = - 2 \times 13.6 \times \frac{z^{2}}{n^{2}} P E i n {H e}^{+} = - 2 \times 13.6 \times 4 = - 108.8 e V {P E}_{H} > {P E}_{{H e}^{+}}$
KE in H= $13.6 \times \frac{z^{2}}{n^{2}} = 13.6 e V (z = 1, n = 1)$
$K E = 13.6 \times \frac{z^{2}}{n^{2}}$
KE in ${H e}^{+} = 13.6 \times \frac{z^{2}}{n^{2}} = 54.4 e V (z = 2, n = 2)$
Thus KE is more in $H e^{+}$

Suggest Corrections

Similar questions

H e^{+}

E_{P}

E_{K}

\frac{E_{p}}{E_{K}}

In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is

1 s t

2.18 \times 10^{- 18} J

5 t h

Join BYJU'S Learning Program