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(C) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{

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$n_2=2$ to $n_1=1$

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(C) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ $(\lambda = \frac{hc}{\Delta E})$.To obtain the lowest wavelength,we need the transition with the highest energy difference $\Delta E$.Comparing the transitions:$(A)$ $n_2=\infty$ to $n_1=2$: $\Delta E \propto (1/4 - 0) = 0.25$$(B)$ $n_2=4$ to $n_1=3$: $\Delta E \propto (1/9 - 1/16) = 0.0486$$(C)$ $n_2=2$ to $n_1=1$: $\Delta E \propto (1/1 - 1/4) = 0.75$$(D)$ $n_2=5$ to $n_1=3$: $\Delta E \propto (1/9 - 1/25) = 0.0711$The transition $n_2=2$ to $n_1=1$ has the largest energy difference,therefore it emits radiation of the lowest wavelength.

Which one of the following transitions of an electron in a hydrogen atom emits radiation of the lowest wavelength?

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