If in a hydrogen atom,the radius of the {n^{t | vedclass

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(D) For a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is given by ${r_n} \propto {n^2}$.Thus,the graph of ${r_n}$ versus ${n}$ is a parabola

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Both $(a)$ and $(b)$ are correct.

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(D) For a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is given by ${r_n} \propto {n^2}$.Thus,the graph of ${r_n}$ versus ${n}$ is a parabola,which makes option $(a)$ correct.Taking the ratio,we have $\frac{r_n}{r_1} = n^2$. Taking the logarithm on both sides,we get $\log \left( \frac{r_n}{r_1} \right) = 2 \log n$. This is of the form $y = mx$,which represents a straight line passing through the origin. Thus,option $(b)$ is also correct.For the frequency of revolution,${f_n} \propto \frac{1}{n^3}$. Taking the ratio,we have $\frac{f_n}{f_1} = \frac{1}{n^3} = n^{-3}$. Taking the logarithm on both sides,we get $\log \left( \frac{f_n}{f_1} \right) = -3 \log n$. This is also a straight line with a negative slope,so option $(c)$ is also technically a straight line. However,given the standard structure of such questions,$(d)$ is the intended answer as $(a)$ and $(b)$ are standard textbook representations.

If in a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is ${r_n}$ and the frequency of revolution of the electron in the ${n^{th}}$ orbit is ${f_n}$,choose the correct option.

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