Question 120 of 239

The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is

A$D$B$A$C$B$D$C$

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Answer: (A) $D$

Wavelength of photons released during de-excitation is given by,
$\dfrac{1}{\lambda }=R\left[\dfrac{1}{{\left({n}_{1}\right)}^{2}}-\dfrac{1}{{\left({n}_{2}\right)}^{2}}\right]$.
Clearly, shortest wavelength will correspond to highest energy difference in de-excitation, which in the given figure will correspond to transition from $n=3$ to $n=1.$

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