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The radius of the first Bohr orbit of ${He}^{+}$ having $n = 1$ is about (Given: $e^{2 /} 4 π ε_{0} = 2.30 \times 10^{- 28} {Nm}^{2}$ )

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$53.4 pm$

$106.8 pm$

$26.7 pm$

$42.1 pm$

answer is C.

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Detailed Solution

$r = n^{2} [\frac{h^{2}}{4 π^{2} m (Z e^{2} / 4 π ε_{0})}] = (1)^{2} [\frac{{(6.626 \times 10^{- 34} Js)}^{2}}{4 (3.14)^{2} (9.1 \times 10^{- 31} kg) (2 \times 2.30 \times 10^{- 28} {Nm}^{2})}] +$

$= 2.66 \times 10^{- 11} m = 26.6 pm$

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