A light of energy 12.75 ; eV is incident on a | vedclass

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(B) The energy of the hydrogen atom in the ground state $(n=1)$ is $E_1 = -13.6 \; eV$.When the atom absorbs a photon of energy $12.75 \; eV$,its new

Vedclass · Accepted Answer

$828$

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(B) The energy of the hydrogen atom in the ground state $(n=1)$ is $E_1 = -13.6 \; eV$.When the atom absorbs a photon of energy $12.75 \; eV$,its new energy $E_n$ becomes:$E_n = E_1 + 12.75 = -13.6 + 12.75 = -0.85 \; eV$.Using the formula $E_n = \frac{-13.6}{n^2} \; eV$,we have:$\frac{-13.6}{n^2} = -0.85$$n^2 = \frac{13.6}{0.85} = 16$$n = 4$.The angular momentum $L$ of an electron in the $n$-th orbit is given by Bohr's postulate:$L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$.Substituting $h = 4.14 	imes 10^{-15} \; eVs$:$L = \frac{2 	imes 4.14 	imes 10^{-15}}{\pi} = \frac{8.28 	imes 10^{-15}}{\pi} = \frac{828 	imes 10^{-17}}{\pi} \; eVs$.Comparing this with $\frac{x}{\pi} 	imes 10^{-17} \; eVs$,we get $x = 828$.

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