Q147 / 200 · WBJEE 2013 · Physics
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$+3.4 \mathrm{eV}$
- B$-3.4 \mathrm{eV}$
- C$+6.8 \mathrm{eV}$
- Correct answerD$-6.8 \mathrm{eV}$
Solution
Given, $E_{n}=13.6 \mathrm{eV}$
Energy of an electron in $n^{\text {th }}$ state
$E_{n}=\frac{-13.6 z^{2} \mathrm{eV}}{n^{2}}$
$\therefore$ Energy of an electron in $n=2$ state
$E_{2}=\frac{18-6 z^{2}}{(2)^{2}}=-34 e V$
So.
$\begin{aligned}<br />P E &=2 E_{n=2} \\<br />&=2 \times(-3.4) \\<br />&=-6.8 \mathrm{eV}<br />\end{aligned}$
Energy of an electron in $n^{\text {th }}$ state
$E_{n}=\frac{-13.6 z^{2} \mathrm{eV}}{n^{2}}$
$\therefore$ Energy of an electron in $n=2$ state
$E_{2}=\frac{18-6 z^{2}}{(2)^{2}}=-34 e V$
So.
$\begin{aligned}<br />P E &=2 E_{n=2} \\<br />&=2 \times(-3.4) \\<br />&=-6.8 \mathrm{eV}<br />\end{aligned}$
Hence, the correct answer is option: 'D'.