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(D) In the given hypothetical world, the angular momentum is $L = 2n' \frac{h}{2\pi} = n' \frac{h}{\pi}$, where $n' = 1, 2, 3, \dots$.Comparing this w

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$487$

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(D) In the given hypothetical world, the angular momentum is $L = 2n' \frac{h}{2\pi} = n' \frac{h}{\pi}$, where $n' = 1, 2, 3, \dots$.Comparing this with the standard Bohr quantization $L = n \frac{h}{2\pi}$, we see that the allowed orbits correspond to $n = 2n'$.The energy of an orbit is $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. Substituting $n = 2n'$, we get $E_{n'} = -\frac{13.6}{(2n')^2} = -\frac{13.6}{4n'^2} = -\frac{3.4}{n'^2} 	ext{ eV}$.For the visible range, the transition must end at the first excited state of this system. The ground state is $n'=1$ $(n=2)$, and the first excited state is $n'=2$ $(n=4)$.The transition for the largest wavelength (smallest energy) in the visible range is from $n'=2$ to $n'=1$.The energy difference is $\Delta E = E_2 - E_1 = -\frac{3.4}{2^2} - (-\frac{3.4}{1^2}) = -0.85 + 3.4 = 2.55 	ext{ eV}$.The wavelength is $\lambda = \frac{hc}{\Delta E} = \frac{1242 	ext{ eV-nm}}{2.55 	ext{ eV}} \approx 487 	ext{ nm}$.

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