In a hydrogen atom,the radius of the n^{th} B | vedclass

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(A) We know that the radius of the $n^{th}$ Bohr orbit is given by $r_n \propto n^2$.Therefore,we can write $\frac{r_n}{r_1} = n^2$.Taking the logarit

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(A) We know that the radius of the $n^{th}$ Bohr orbit is given by $r_n \propto n^2$.Therefore,we can write $\frac{r_n}{r_1} = n^2$.Taking the logarithm on both sides,we get:$\log \left( \frac{r_n}{r_1} \right) = \log (n^2)$Using the logarithmic property $\log (a^b) = b \log a$,we have:$\log \left( \frac{r_n}{r_1} \right) = 2 \log n$.This equation is of the form $y = mx$,where $y = \log \left( \frac{r_n}{r_1} \right)$,$x = \log n$,and the slope $m = 2$.Since this is a linear equation passing through the origin with a positive slope,the graph will be a straight line passing through the origin.

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

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