Question

If kinetic energy of electron revolving in first orbit of $H$-atom is $E,$ then potential energy of electron revolving in second orbit of $H{e}^{+}$
(correct answer + 1, wrong answer - 0.25)

• A. $E$
• B. $-E$
• C. $+2E$
• D. $-2E$

Answer 1

The correct option is D $-2E$

As we know that the kinetic energy of electron revolution in first orbit of $H$-atom is given by
$E=+13.6\text{eV}$

For $H{e}^{+}$,
$U=2[\frac{-Z^2}{n^2}\times 13.6\text{eV}]\Rightarrow U=2[\frac{-4}{4}\times 13.6\text{eV}]$
$=-2\times 13.6\text{eV}=-2\text{E}$