Question

The energy of electron in first Bohr’s orbit of $H$-atom is $-13.6eV$. What will be its potential energy in $n=4^{th}$ orbit.

• A. $-13.6eV$
• B. $-3.4eV$
• C. $-0.85eV$
• D. $-1.70eV$

Answer 1

Answer: (D)

Energy of nth orbit of H-atom
$=-\frac{2{\pi }^{2}m{e}^4{k}^2}{h^2}\times \frac{1}{n^2}$
Energy of nth orbit of H-atom
$=-\frac{2{\pi }^{2}m{e}^4{k}^2}{h^2}$
$=-13.6ev\left(given\right)$
Energy of fourth Bohr’s orbit of H-atom $=-\frac{2{\pi }^{2}m{e}^4{k}^2}{h^2}\times \frac{1}{4^2}$
$=13.6\times \frac{1}{16}eV=-0.85ev$
PE of electron in nth orbit $=2\times E_n$
So P.E. of electron in 4th orbit $=2\times \left(-0.85\right)=-1.70eV$