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

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decreases

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.The energy difference between adjacent levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = 13.6 	imes Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight) = 13.6 	imes Z^2 \left( \frac{2n+1}{n^2(n+1)^2} ight)$.As $n$ increases,the term $\frac{2n+1}{n^2(n+1)^2}$ decreases rapidly.For example:For $n=1$ to $2$,$\Delta E = 13.6 	imes (1 - 0.25) = 10.2 	ext{ eV}$.For $n=2$ to $3$,$\Delta E = 13.6 	imes (0.25 - 0.111) = 1.89 	ext{ eV}$.Thus,the energy difference decreases as the principal quantum number increases.

With increasing principal quantum number,the energy difference between adjacent energy levels in $H$ atom

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