Angular velocity of electron in the ground state of Hydrogen atom is $ω_{0}$ .Then what will be the angular velocity of electron in the second excited state of ${He}^{+}$ atom?

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$\frac{4 ω_{0}}{27}$

$\frac{9 ω_{0}}{16}$

$\frac{3 ω_{0}}{4}$

$\frac{2 ω_{0}}{9}$

answer is C.

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Detailed Solution

$mvr = {mωr}^{2} = n . \frac{h}{2 π} \Rightarrow ω \propto \frac{n}{r^{2}}$

$Also, r = r_{0} . \frac{n^{2}}{z} combining, ω \propto \frac{z^{2}}{n^{3}}$

$∴ \frac{ω_{He}}{ω_{H}} = {(\frac{2}{1})}^{2} x \frac{1^{3}}{3^{3}} \Rightarrow ω_{He} = \frac{4 ω_{0}}{27}$

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