Question

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

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

Answer 1

Answer: (D)

$-6.8eV$

Energy of the nth state of hydrogen atom $E_n=\frac{-13.6}{n^2}eV$

Thus energy of $n=2$ state $E_2=\frac{-13.6}{2^2}=-3.4eV$

Potential energy of electron in $n=2$ state of hydrogen atom $U=2E$

$⟹$ $U=2\times (-3.4)=-6.8eV$