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(B) The frequency of orbital motion $f$ of an electron in a hydrogen-like atom is given by $f \propto \frac{Z^2}{n^3}$.Since the atom is hydrogen,$Z =

Vedclass · Accepted Answer

$n_1 = 3, n_2 = 1$

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(B) The frequency of orbital motion $f$ of an electron in a hydrogen-like atom is given by $f \propto \frac{Z^2}{n^3}$.Since the atom is hydrogen,$Z = 1$,so $f \propto \frac{1}{n^3}$.Let $f_1$ be the frequency in the initial state $n_1$ and $f_2$ be the frequency in the final state $n_2$.Given that $f_1 = \frac{1}{27} f_2$,we have $\frac{1}{n_1^3} = \frac{1}{27} \cdot \frac{1}{n_2^3}$.Rearranging the terms,we get $\frac{n_2^3}{n_1^3} = \frac{1}{27}$.Taking the cube root of both sides,$\frac{n_2}{n_1} = \frac{1}{3}$,which implies $n_1 = 3n_2$.Checking the options:For option $B$,$n_1 = 3$ and $n_2 = 1$,which satisfies $n_1 = 3(1) = 3$.Thus,the correct values are $n_1 = 3$ and $n_2 = 1$.

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