The energy required (in eV) to excite an electron of $\mathrm{H}$-atom from the ground state to the third state is

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Aicharya · Accepted Answer

$E=\frac{-13.6 Z^2}{n^2}$ $E_1=\frac{-13.6 	imes 1^2}{1^2}$ $=-13.6 \mathrm{eV}$ $E_2=\frac{-13.6 	imes 1^2}{3^2}$ $=-1.5111 \mathrm{eV}$ $	herefore \quad \Delta E=E_2-E_1$ $\begin{aligned} & =-1.511-(-13.6) \ & =12.08 \mathrm{eV} \ & \cong 12.1 \mathrm{eV}\end{aligned}$

The energy required (in eV) to excite an electron of $\mathrm{H}$-atom from the ground state to the third…

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