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(C) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $E = hc / \lambda$. The energy is least when the transition is

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$122$

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(C) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $E = hc / \lambda$. The energy is least when the transition is between the closest energy levels.For the Lyman series, the transitions occur to $n_1 = 1$. The least energetic photon corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.The energy difference is $\Delta E = 13.6 	ext{ eV} 	imes (1/n_1^2 - 1/n_2^2) = 13.6 	imes (1/1^2 - 1/2^2) = 13.6 	imes (3/4) = 10.2 	ext{ eV}$.Using the relation $\lambda = hc / \Delta E$:$\lambda = 1240 	ext{ eV nm} / 10.2 	ext{ eV} \approx 121.57 	ext{ nm} \approx 122 	ext{ nm}$.

What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum (in $\text{ nm}$)? (Take $hc = 1240 \text{ eV nm}$)

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