Question

The kinetic energy of electron in the first Bohr orbit will be:

• A. $13.6eV$
• B. $-489.6eV$
• C. $0.38eV$
• D. $-0.38eV$

Answer 1

The correct option is A $13.6eV$

The kinetic energy of an electron in a hydrogen atom is:
$KE=\frac{m{e}^4}{8n^2{h}^2{ϵ}_0^2}$

For $n=1$,
$KE=\frac{m{e}^4}{8n^2{h}^2{ϵ}_0^2}$
$KE=\frac{9.1\times {10}^{-31}(1.6\times {10}^{-16}{)}^4}{8\left(1{)}^2\right(6.6\times {10}^{-34}{)}^2(8.85\times {10}^{-12}{)}^2}$

$KE=\frac{59.63\times {10}^{-107}}{27293.82\times {10}^{-92}}$

$KE=0.002185\times {10}^{-15}$

$KE=\frac{0.002185\times {10}^{-15}}{1.6\times {10}^{-19}}eV$

$KE=0.001365\times {10}^{4}eV$

$KE=13.6eV$