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:

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$5.3 \times 10^{- 11} m$

$4.3 \times 10^{- 11} m$

$6.3 \times 10^{- 11} m$

$7.3 \times 10^{- 11} m$

answer is A.

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Detailed Solution

Here, $E = - 13.6 e V = - 13.6 \times 1.6 \times 10^{- 19} = - 2.2 \times 10^{- 18} J$

$E = \frac{- e^{2}}{8 π \in_{0} r}$

$∴$ As orbital radius,

$r = \frac{- e^{2}}{8 π \in_{0} r} = \frac{9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{2 \times (2.2 \times 10^{- 18})} = 5.3 \times 10^{- 11} m$ .

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