In hydrogen atom radius of the first orbit is $0.53 \overset{O}{A}$ . Minimum work required to free the electron, given its K.E in the orbit is half the magnitude of P.E in the orbit.

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8.6eV

6.8eV

27.2eV

13.6eV

answer is A.

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Detailed Solution

To free electron , $n_{1} = 1; n_{2} = \infty; W = ΔE$
$= {TE}_{f} - {TE}_{i} = (\frac{- 13 .6 (1)^{2}}{\infty^{2}}) - (\frac{- 13 .6 (1)^{2}}{1^{2}})$
= 0 – (–13.6) = 13.6eV

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