Which of the following transitions in hydrogen atom emit photons of highest frequency? ( n = principle quantum number)

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Emission of a photon occurs when an electron transitions from a higher energy level to a lower energy level. Thus, transitions $n = 1$ to $n = 3$ and $n = 2$ to $n = 4$ represent absorption of energy, not emission.

For the emission transitions, the energy of the emitted photon is given by $Δ E = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV$ .

For $n = 5$ to $n = 3$ :
$Δ E = 13.6 (\frac{1}{3 ^{2}} - \frac{1}{5 ^{2}}) = 13.6 (\frac{16}{225}) \approx 0.97 eV$

For $n = 2$ to $n = 1$ :
$Δ E = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}) = 13.6 (\frac{3}{4}) = 10.2 eV$

Since frequency $ν = \frac{Δ E}{h}$ , the transition with the highest energy difference emits the photon of the highest frequency. Therefore, the transition $n = 2$ to $n = 1$ emits photons of the highest frequency.

Answer: $n = 2$ to $n = 1$

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