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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.When an electron jumps from an initi

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.When an electron jumps from an initial orbit $n_1 = 2$ to a final orbit $n_2 = 1$,the energy emitted is given by the difference in energy levels:$\Delta E = E_{n_1} - E_{n_2} = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ (for emitted energy,we take the magnitude of the change).$\Delta E = -13.6 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) \ eV$$\Delta E = -13.6 \left( \frac{1}{4} - 1 \right) \ eV$$\Delta E = -13.6 \left( -\frac{3}{4} \right) \ eV$$\Delta E = 10.2 \ eV$.Since the question asks for the energy emitted,the value is $10.2 \ eV$. None of the provided options match this result.

In a hydrogen atom,when an electron jumps from the second to the first orbit,the energy emitted is:

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