Which of the following electronic transitions | vedclass

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(C) The wavelength $(\lambda)$ of an emitted photon is inversely proportional to the energy difference $(\Delta E)$ between the two energy levels,give

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$n = 4 \rightarrow n = 1$

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(C) The wavelength $(\lambda)$ of an emitted photon is inversely proportional to the energy difference $(\Delta E)$ between the two energy levels,given by the relation $\Delta E = \frac{hc}{\lambda}$.To obtain the shortest wavelength,the energy difference $(\Delta E)$ must be the maximum.The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ 	ext{eV} / n^2$.The energy difference for a transition from $n_2$ to $n_1$ is $\Delta E = 13.6 \ 	ext{eV} 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Comparing the given transitions:$(A)$ $n = 2 ightarrow n = 1$: $\Delta E \propto (1 - 0.25) = 0.75$$(B)$ $n = 3 ightarrow n = 1$: $\Delta E \propto (1 - 0.111) = 0.889$$(C)$ $n = 4 ightarrow n = 1$: $\Delta E \propto (1 - 0.0625) = 0.9375$$(D)$ $n = 4 ightarrow n = 3$: $\Delta E \propto (0.111 - 0.0625) = 0.0485$The transition $n = 4 ightarrow n = 1$ has the largest energy difference,and therefore,it corresponds to the shortest wavelength.

Which of the following electronic transitions in a hydrogen atom results in the emission of a spectral line with the shortest wavelength?

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