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JEE Main 2018 Lyman Pfund Series Problem

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If the series limit frequency of the Lyman series is \nu_L , then the series limit frequency of the Pfund series is:

(1) 25\nu_L
(2) 16\nu_L
(3) \frac{\nu _L}{16}
(4) \frac{\nu_L}{25}

[Based on JEE Main 2018]

Solutions

(A) Text Format Solution

We have, \frac {1}{\lambda} \propto \frac {1}{n_1 ^2}-\frac {1}{n_2 ^2}

Or \nu \propto \frac {1}{n_1 ^2}-\frac {1}{n_2 ^2}

In case of series limit, n_2 \to \infty and therefore \nu \propto \frac {1}{n_1 ^2}

For Lyman Series, \nu_L \propto \frac {1}{1^2}

For Pfund Series, \nu_P \propto \frac {1}{5^2}

\therefore \nu_P=\frac {\nu_L}{25}

Hence, Option (4).

(B) Image Format Solution

If the series limit frequency of the Lyman series is nu_L, then the series limit frequency of the Pfund series is:

(C) .pdf Format Solution

JEE-Main-2018-Lyman-Pfund-Series-Problem.pdf