15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T516,

Lyman series corresponds to electron transitions from higher energy levels ($n_2$) to the first energy level ($n_1 = 1$) in a hydrogen atom.

Step 1: Wavelength formula for Lyman series

The wavelength $\lambda$ is given by:

$ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_2^2} \right) $

where $R$ is the Rydberg constant, $n_1 = 1$, and $n_2 = 2, 3, 4, ...$

Step 2: Minimum wavelength ($P$) of Lyman series

Minimum wavelength occurs when the denominator is the largest, that is, when $n_2 \to \infty$:

$ \frac{1}{P} = R \left(1 - \frac{1}{\infty^2}\right) = R(1 - 0) = R \implies P = \frac{1}{R} $

Step 3: Maximum wavelength of Lyman series

Maximum wavelength occurs when $n_2$ is the smallest possible ($n_2 = 2$):

$ \frac{1}{\lambda_{max}} = R \left(1 - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = R\left(\frac{3}{4}\right) $

$ \lambda_{max} = \frac{1}{R} \cdot \frac{4}{3} = \frac{4}{3} \cdot \frac{1}{R} $

But from Step 2, $P = \frac{1}{R}$. So:

$ \lambda_{max} = \frac{4}{3} P $

Step 4: Answer

Correct option is:

$ \boxed{\frac{4 P}{3}} $

Which is Option A.

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