Question

The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6eV$. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is/are:

• A. $-3.4eV$
• B. $-4.2eV$
• C. $-6.8eV$
• D. $-1.5eV$

Answer 1

Answer: (A), (D)

$-3.4eV$
$-1.5eV$

Energy in an orbit of hydrogen atom, $E_n$ is given by

$E_n=\frac{2{\pi }^{2}m{e}^4{Z}^2}{n^2{h}^2}=\frac{constant}{n^2}=\frac{-13.6eV}{n^2}$
For $1^{st}$ orbit, n=1, hence $E_1=\frac{-13.6eV}{1}=-13.6eV$
For $2^{nd}$ orbit, n=2, hence $E_2=\frac{-13.6eV}{4}=-3.4eV$
For $3^{rd}$ orbit, n=3, hence $E_3=\frac{-13.6eV}{9}=-1.51eV$
For $4^{th}$ orbit, n=4, hence $E_4=\frac{-13.6eV}{16}=-0.85eV$