Question

The energy of an electron in the $n^{th}$ Bohr orbit of hydrogen atom is:

• A. $\frac{13.6}{n^3}eV$
• B. $-\frac{13.6}{n^3}eV$
• C. $-\frac{13.6}{n^2}eV$
• D. $-\frac{13.6}{n}eV$

Answer 1

The correct option is C $-\frac{13.6}{n^2}eV$

According to the problem:

Let us consider the Bohr's orbit of H atom is gen by the expression

$E_n=\frac{2{\pi }^{2}m{e}^4{Z}^2}{n^2{h}^2{(4\pi \in 0)}^2}$

implies that

$=13.6\frac{Z^2}{v^2}eV$

$(SinceZ=1)$

Therefore,

$E_n=\frac{-13.6}{n^2}eV$.

Hence, the correct option is $\text{C}$