With the increase in principal quantum number | vedclass

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy difference between

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Decreases

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(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy difference between two successive levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = -13.6 \left( \frac{1}{(n+1)^2} - \frac{1}{n^2} ight) = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight) = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} ight) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} ight)$.As the principal quantum number $n$ increases,the denominator $n^2(n+1)^2$ increases much faster than the numerator $(2n+1)$.Therefore,the energy difference $\Delta E$ decreases as $n$ increases.From the provided image,we can see that the energy gap between levels decreases as we move to higher energy states: $E_1 > E_2 > E_3$.

With the increase in principal quantum number,the energy difference between two successive energy levels

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