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

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$2$

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(D) The energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.The fourth excited state corresponds to $n_1 = 5$ (since ground state is $n=1$,first excited is $n=2$,...,fourth excited is $n=5$).The energy emitted during a transition from $n_1$ to $n$ is given by $\Delta E = 13.6 \left( \frac{1}{n^2} - \frac{1}{n_1^2} ight) \ eV$.Given $\Delta E = 2.86 \ eV$ and $n_1 = 5$,we have:$2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} ight)$$\frac{2.86}{13.6} = \frac{1}{n^2} - \frac{1}{25}$$0.2103 \approx \frac{1}{n^2} - 0.04$$\frac{1}{n^2} = 0.2103 + 0.04 = 0.2503 \approx \frac{1}{4}$$n^2 = 4 \implies n = 2$.

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy $2.86 \ eV$. The integer value of $n$ will be . . . . . . .

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