In a hydrogen atom,two electrons are in orbit | vedclass

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(C) According to Bohr's model,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$. Given $r_1 = r_0$ and $r_2 = 4r_0$,we have $n_1^2 \propt

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$8:1$

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(C) According to Bohr's model,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$. Given $r_1 = r_0$ and $r_2 = 4r_0$,we have $n_1^2 \propto r_0$ and $n_2^2 \propto 4r_0$. Thus,$n_2/n_1 = \sqrt{4r_0/r_0} = 2$. The velocity of an electron in the $n^{th}$ orbit is $v_n \propto 1/n$. The frequency of revolution $f$ is given by $f = v / (2\pi r)$. Substituting $v \propto 1/n$ and $r \propto n^2$,we get $f \propto (1/n) / n^2 = 1/n^3$. Therefore,the ratio of frequencies is $f_1/f_2 = (n_2/n_1)^3 = (2)^3 = 8/1$. Thus,the ratio is $8:1$.

In a hydrogen atom,two electrons are in orbits of radii $r_0$ and $4r_0$. What is the ratio of their frequencies of revolution around the nucleus?

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