As the n (number of orbit) increases,the diff | vedclass

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

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Decreases

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.The energy difference between two consecutive orbits $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = -13.6 \left( \frac{1}{(n+1)^2} - \frac{1}{n^2} \right) = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.As $n$ increases,the denominator $n^2(n+1)^2$ grows much faster than the numerator $(2n+1)$,causing the value of $\Delta E$ to decrease.Therefore,the energy difference between consecutive energy levels decreases as $n$ increases.

As the $n$ (number of orbit) increases,the difference of energy between the consecutive energy levels

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