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(A) The area enclosed by the $n^{th}$ orbit is given by $A_n = \pi r_n^2$.Since the radius of the $n^{th}$ orbit in a hydrogen-like atom is proportion

Vedclass · Accepted Answer

$4$

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(A) The area enclosed by the $n^{th}$ orbit is given by $A_n = \pi r_n^2$.Since the radius of the $n^{th}$ orbit in a hydrogen-like atom is proportional to $n^2$ (i.e., $r_n \propto n^2$), we have:$\frac{A_n}{A_1} = \left( \frac{r_n}{r_1} \right)^2 = \left( \frac{n^2}{1^2} \right)^2 = n^4$.Taking the natural logarithm $(\ln)$ on both sides:$\ln \left( \frac{A_n}{A_1} \right) = \ln (n^4) = 4 \ln (n)$.Comparing this with the equation of a straight line $y = mx + c$, where $y = \ln \left| \frac{A_n}{A_1} \right|$, $x = \ln |n|$, $m = 4$, and $c = 0$, we see that the graph is a straight line passing through the origin with a slope of $4$.Looking at the provided graph, curve $4$ represents a straight line with a slope of $4$ (since at $\ln |n| = 1$, $\ln \left| \frac{A_n}{A_1} \right| = 4$).Therefore, the correct curve is $4$.

The figure shows a graph between $\ln \left| \frac{A_n}{A_1} \right|$ and $\ln |n|$, where $A_n$ is the area enclosed by the $n^{th}$ orbit in a hydrogen-like atom. The correct curve is

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