Question

If the shortest wavelength in Lyman series of hydrogen atom is $A$, then the longest wavelength in Paschen series of $He^{+}$ is :

• A. $\frac{5A}{9}$
• B. $\frac{9A}{5}$
• C. $\frac{36A}{5}$
• D. $\frac{36A}{7}$

Answer 1

The Correct Option is D

To solve this problem, we need to understand the transitions in different spectral series of hydrogen-like atoms. The Lyman series involves transitions to the first energy level (n=1), and the Paschen series involves transitions to the third energy level (n=3). The question involves both the hydrogen atom and the helium ion (\(He^+\)). Let's break down the problem step-by-step:

Step 1: Lyman Series of Hydrogen Atom

The shortest wavelength in the Lyman series corresponds to the transition from \(n=\infty\) to \(n=1\).

The formula for the wavelength of emitted light during an electronic transition in a hydrogen-like atom is given by:

\[\frac{1}{\lambda} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]

For the hydrogen atom (\(Z=1\)) transitioning from \(n=\infty\) to \(n=1\), the shortest wavelength is:

\[\frac{1}{A} = R \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R\]

Thus, \(A = \frac{1}{R}\).

Step 2: Paschen Series of \(He^+\)

The longest wavelength in the Paschen series corresponds to the transition from \(n=4\) to \(n=3\), since the energy difference is smallest.

For the \(He^+\) ion, \(Z=2\). Using the same formula:

\[\frac{1}{\lambda} = R(2)^2 \left(\frac{1}{3^2} - \frac{1}{4^2}\right)\]

This simplifies to:

\[\frac{1}{\lambda} = 4R \left(\frac{1}{9} - \frac{1}{16}\right) = 4R \left(\frac{7}{144}\right) = \frac{28R}{144} = \frac{7R}{36}\]

Therefore, the longest wavelength \(\lambda\) is:

\[\lambda = \frac{36}{7R}\]

Step 3: Relate to Given \(A\)

Since \(A = \frac{1}{R}\), substitute \(R = \frac{1}{A}\) in the wavelength formula for \(\lambda\) we derived:

\[\lambda = \frac{36}{7} \times A\]

Thus, the longest wavelength in the Paschen series of \(He^+\) is \(\frac{36A}{7}\).

Conclusion

The correct answer is \(\frac{36A}{7}\), matching the provided correct option.