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. $+ 3.4\, eV$
• B. $- 3.4\, eV$
• C. $+ 6.8\, eV$
• D. $- 6.8\, eV$

Answer 1

Answer: (D)

$- 6.8\, eV$

Answer 2

Given, $E_{n}=13.6\, e V$
Energy of an electron in $n^{\text {th }}$ state
$E_{n}=\frac{-13.6 z^{2} e V}{n^{2}}$
$\therefore$ Energy of an electron in $n=2$ state So,
$E_{2}= \frac{18-6 z^{2}}{(2)^{2}}=-3.4 \,e V$
$P E =2 E_{n=2}$
$=2 \times(-3.4)$
$\approx-6.8 \,eV$