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The speed of an electron in the innermost orbit of the hydrogen atom (Bohr radius = 52.9 pm, $m_{e} = 9.11 \times 10^{- 31} kg)$ is

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2.19 \times 10^{4} {ms}^{- 1}

2.19 \times 10^{6} {ms}^{- 1}

2.19 \times 10^{7} {ms}^{- 1}

2.19 \times 10^{8} {ms}^{- 1}

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detailed solution

Correct option is B

The required speed is

$v = \frac{2 π}{h} (\frac{e^{2}}{4 π ε_{0}}) = \frac{2 (3.14)}{(6.626 \times 10^{- 34} Js)} [\frac{{(1.602 \times 10^{- 19} C)}^{2}}{1.112 \times 10^{- 10} C^{2} N^{- 1} m^{- 2}}] = 2.19 \times 10^{6} {ms}^{- 1}$

The option B is correct.

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The speed of an electron in the innermost orbit of the hydrogen atom (Bohr radius = 52.9 pm, me=9.11×10−31kg is

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The speed of an electron in the innermost orbit of the hydrogen atom (Bohr radius = 52.9 pm, $m_{e} = 9.11 \times 10^{- 31} kg)$ is