As the quantum number increases, the difference in energy between consecutive energy levels

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Aicharya · Accepted Answer

$E_{n+1}-E_n=13.6\left[\frac{1}{n^2}-\frac{1}{(n+1)^2}ight]$ $\begin{aligned} & \Delta \mathrm{E}_{\mathrm{n}}=13.6\left[\frac{(\mathrm{n}+1)^2-\mathrm{n}^2}{\mathrm{n}^2(\mathrm{n}+1)^2}ight] \ & =13.6\left[\frac{1+2 \mathrm{n}}{\mathrm{n}^2(\mathrm{n}+1)^2}ight]\end{aligned}$ For large value of $n$ $\Delta \mathrm{E}_{\mathrm{n}}=13.6\left[\frac{2 \mathrm{n}}{\mathrm{n}^4}ight]$ $\begin{aligned} & \propto \frac{1}{\mathrm{n}^3} \ & 	ext { So, } \mathrm{n} \uparrow \uparrow \Rightarrow \Delta \mathrm{E}_{\mathrm{n}} \downarrow \downarrow\end{aligned}$

As the quantum number increases, the difference in energy between consecutive energy levels

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