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Maximum frequency of emission is obtained for the transition :
A
n = 2 to n = 1
B
n = 6 to n = 2
C
n = 1 to n = 2
D
n = 2 to n = 6
Detailed Solution
Emission occurs only when an electron falls from a higher to a lower level; n = 1 to 2 and n = 2 to 6 are absorptions.
Frequency $\nu \propto \Delta E = 13.6\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV
n = 2 to 1: $\Delta E = 13.6\left(1 - \dfrac{1}{4}\right) = 10.2$ eV
n = 6 to 2: $\Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{36}\right) = 3.02$ eV
So the transition n = 2 to n = 1 gives the maximum frequency of emission.
Frequency $\nu \propto \Delta E = 13.6\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV
n = 2 to 1: $\Delta E = 13.6\left(1 - \dfrac{1}{4}\right) = 10.2$ eV
n = 6 to 2: $\Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{36}\right) = 3.02$ eV
So the transition n = 2 to n = 1 gives the maximum frequency of emission.