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According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
A
n = 6 to n = 5
B
n = 5 to n = 3
C
n = 6 to n = 1
D
n = 5 to n = 4
Detailed Solution
Energy of the photon emitted: $\Delta E = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV, where $n_2 > n_1$.
n = 6 to n = 5: $13.6\left(\frac{1}{25} - \frac{1}{36}\right) = 13.6\times0.0122 = 0.166$ eV
n = 5 to n = 4: $13.6\left(\frac{1}{16} - \frac{1}{25}\right) = 13.6\times0.0225 = 0.306$ eV
n = 5 to n = 3: $13.6\left(\frac{1}{9} - \frac{1}{25}\right) = 13.6\times0.0711 = 0.967$ eV
n = 6 to n = 1: $13.6\left(1 - \frac{1}{36}\right) = 13.6\times0.972 = 13.22$ eV
The energy levels come closer together as n increases, so the transition between the highest adjacent levels gives the smallest energy.
The least energetic photon is from n = 6 to n = 5.
n = 6 to n = 5: $13.6\left(\frac{1}{25} - \frac{1}{36}\right) = 13.6\times0.0122 = 0.166$ eV
n = 5 to n = 4: $13.6\left(\frac{1}{16} - \frac{1}{25}\right) = 13.6\times0.0225 = 0.306$ eV
n = 5 to n = 3: $13.6\left(\frac{1}{9} - \frac{1}{25}\right) = 13.6\times0.0711 = 0.967$ eV
n = 6 to n = 1: $13.6\left(1 - \frac{1}{36}\right) = 13.6\times0.972 = 13.22$ eV
The energy levels come closer together as n increases, so the transition between the highest adjacent levels gives the smallest energy.
The least energetic photon is from n = 6 to n = 5.