Question

The ground state energy of hydrogen atom is $-13.6eV$. What is the potential energy of the electron in this state?

• A. 0 eV
• B. −27.2 eV
• C. 1 eV
• D. 2 eV

Answer 1

Answer: (B)

From the first postulate of Bohr's atom model.
$\frac{m{v}^2}{r}=\frac{KZ{e}^2}{r^2}$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}\frac{KZ{e}^2}{r}$
i.e., $KE$ of electron $=\frac{1}{2}m{v}^2=\frac{KZ{e}^2}{2r}$ ...(1)
Potential due to the nucleus, in the orbit in which electron is revolving
$=\frac{KZe}{r}$
$\therefore$ Potential energy of electron
potential $\times$ charge
$=\frac{KZe}{r}(-e)=-\frac{KZ{e}^2}{r}$ ...(2)
Total energy of electron in the orbit
$E=KE+PE$
$=\frac{1}{2}\frac{KZ{e}^2}{r}-\frac{KZ{e}^2}{r}$
$=-\frac{KZ{e}^2}{2r}$ ...(3)
This implies that
$PE=2E$
Given, that in ground state of hydrogen, total energy
$E=-13.6eV$
$PE=-2\times 13.6$
Total energy of electron in outer orbits is more than