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In Bohr model of hydrogen atom, magnitude of force acting on the electron in the ground state of hydrogen atom is F. Then the magnitude of force acting on the electron of He⁺ ion in the second excited state is

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$\frac{4 F}{27}$

$\frac{8 F}{81}$

$\frac{16 F}{9}$

$\frac{8 F}{27}$

answer is B.

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Detailed Solution

In the nth energy state in a hydrogen like ion, $V_{n} \propto \frac{z}{n} and r_{n} \propto \frac{n^{2}}{z}$

$∴ F_{n} - α \frac{V_{n}^{2}}{r_{n}} \Rightarrow F_{n} \propto \frac{{(\frac{Z}{n})}^{2}}{n^{2} / z} \Rightarrow F_{n} \propto \frac{z^{3}}{n^{4}}$

$∴ \frac{F^{1}}{F} = {(\frac{Z^{1}}{Z})}^{3} \times {(\frac{n}{n^{1}})}^{4} = {(\frac{2}{1})}^{3} . {(\frac{1}{3})}^{4} = \frac{8}{81}$

$\Rightarrow F^{1} = \frac{8 F}{81}$

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