Question

The ionization energy of the hydrogen atom is 13.6 eV. 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)

$-6.8eV$

Total energy of electron in $n=2$ state $E_n=\frac{-13.6}{n^2}$ where $n=2$

$\therefore$ $E_2=\frac{-13.6}{2^2}=-3.4eV$

Thus potential energy of electron $U_2=2E_2$

$⟹$ $U_2=2(-3.4)=-6.8eV$