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Q.
If $13.6 \,eV$ energy is required to separate a hydrogen atom into a proton and an electron, then the orbital radius of electron in a hydrogen atom is
Atoms
Solution:
Here, $E = -13.6\, eV $
$= -13.6 \times1.6 \times10^{-19} $
$= -2.2 \times10^{-18} J $
$ E = \frac{-e^{2}}{8\pi \varepsilon_{0}r}$
$ \therefore $ As orbital radius,
$ r= \frac{-e^{2}}{8\pi \varepsilon_{0}E}$
$ = \frac{9\times10^{9} \times\left(1.6\times10^{-19}\right)^{2}}{2\times\left(2.2\times10^{-18}\right)} $
$= 5.3 \times10^{-11} m$.