Question

For $H$-like atoms :

$E_n=-\frac{Z^{2}Rh}{n^2};u_n=\frac{u_{1}Z}{n}$ and $r_n=\frac{r_1\times n^2}{Z};$ where $Rh$ is Rydberg.

What is the potential energy of electron in $2^{nd}$ orbit of $H$-atom?

• A. $-\frac{Rh}{2}$
• B. $-2Rh$
• C. $-4Rh$
• D. $-Rh$

Answer 1

The correct option is A $-\frac{Rh}{2}$

As we know,
Potential energy is twice of total energy

$E_n=-\frac{Z^{2}Rh}{n^2}$

So,

$PE=2\times E_n=-\frac{2\times 1^2\times Rh}{2^2}$ ---- $(n=2;Z=1$ for $H-atom)$

$\therefore PE=-\frac{Rh}{2}$