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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -\frac{Z^2 R_y h c}{n^2}$,where $R_y$ is the Rydberg co

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$1/n^3$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -\frac{Z^2 R_y h c}{n^2}$,where $R_y$ is the Rydberg constant.The energy difference between the levels $n$ and $(n - 1)$ is $\Delta E = E_n - E_{n-1} = Z^2 R_y h c \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} ight)$.Simplifying the expression: $\Delta E = Z^2 R_y h c \left( \frac{n^2 - (n-1)^2}{n^2(n-1)^2} ight) = Z^2 R_y h c \left( \frac{2n - 1}{n^2(n-1)^2} ight)$.Since $n >> 1$,we can approximate $(n-1) \approx n$ and $(2n-1) \approx 2n$.Thus,$\Delta E \approx Z^2 R_y h c \left( \frac{2n}{n^2 \cdot n^2} ight) = \frac{2 Z^2 R_y h c}{n^3}$.Since $\Delta E = h 
u$,the frequency $
u = \frac{\Delta E}{h} \approx \frac{2 Z^2 R_y c}{n^3}$.Therefore,the frequency $
u \propto 1/n^3$.

In a hydrogen-like atom,an electron makes a transition from an energy level with quantum number $n$ to another with quantum number $(n - 1)$. If $n >> 1$,the frequency of the emitted radiation is proportional to:

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