The speed of an electron in the first Bohr orbit is $7.27 \times 10^{5} m s^{- 1}$ . The speed of electron in the $n = 2$ level of ${Li}^{2 +}$ ion will be

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$1.45 \times 10^{6} m s^{- 1}$

$7.27 \times 10^{5} m s^{- 1}$

$1.82 \times 10^{6} m s^{- 1}$

$1.09 \times 10^{6} m s^{- 1}$

answer is B.

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Detailed Solution

Since $v \propto Z / n$ , we will have

$\frac{v ({Li}^{2 +})}{v (H)} = \frac{3 / 2}{1 / 1} \Rightarrow v ({Li}^{+}) = (\frac{3}{2}) v (H) = (\frac{3}{2}) (7.27 \times 10^{5} m s^{- 1}) = 1.09 \times 10^{6} m {s^{- 1}}^{}$

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