Question

The ionization energy of the hydrogen atom is $13.6eV$. The potential energy of the eletron in $n=2$ state of hydrogen atom is

• A. $+3.4eV$
• B. $-3.4eV$
• C. $+6.8eV$
• D. $-6.8eV$

Answer 1

The correct option is D $-6.8eV$

Energy of an electron in $n^{th}$ state:
$\Rightarrow E_n=\frac{-13.6z^{2}eV}{n^2}$

Potential energy of the electron at any state is twice of total energy.
So, $PE=2E_{n=2}$

$\Rightarrow PE=2\times \frac{(-13.6)\times 1^2}{2^2}eV$

$=2\times (-3.4)$

$=-6.8eV$