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.

Answer: (a). Emitted frequency ν ∝ ((1)/(nf²)-(1)/(ni²)); emission needs a jump from higher n to lower n. (c) and (d) are absorptions. (a) 2→1: 1-14 = 0.75; (b) 6→2: 14-136 = 0.22. So the answer is (a) n=2 to n=1.

Maximum frequency of emission is obtained for the transition | NEET 2000 (AIPMT)

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Transitions, Photon Emission & Excitation

Maximum frequency of emission is obtained for the transition

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Answer: (a)

Emitted frequency $\nu \propto \left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ ; emission needs a jump from higher $n$ to lower $n$ .
(c) and (d) are absorptions. (a) $2\to1$ : $1-\frac14 = 0.75$ ; (b) $6\to2$ : $\frac14-\frac1{36} = 0.22$ .
So the answer is (a) $n=2$ to $n=1$ .

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