Question

The shortest wavelength of hydrogen atom in Lyman series is λ. The longest wavelength in Balmer series of He⁺ is

• A. 5/9λ
• B. 5λ/9
• C. 36λ/5
• D. 9λ/5

Hint:

For problems involving spectral series:
• Use the Rydberg formula to calculate wavelengths for hydrogen-like atoms.
• Identify the energy levels (n1, n2) and the corresponding transitions for the given series.
• Simplify using the atomic number Z and the known relationships between series.

Answer 1

The Correct Option is D

 To solve the problem, we begin by understanding the spectral series of the hydrogen atom and extend this understanding to the He⁺ ion.

The Lyman series corresponds to electronic transitions where an electron falls from a higher energy level to the n=1 level. Similarly, the Balmer series corresponds to transitions where an electron falls from a higher energy level to the n=2 level.

The wavelength (\(\lambda\)) of emitted or absorbed light for such transitions can be calculated using the Rydberg formula:

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

Where:

• \(R\) is the Rydberg constant.
• \(Z\) is the atomic number (1 for Hydrogen, 2 for He⁺).
• \(n_1\) and \(n_2\) are principal quantum numbers, with \(n_2 > n_1\).

**Step 1: Calculating shortest wavelength of Lyman series for Hydrogen.**

For hydrogen (Z=1), in the Lyman series where \(n_1 = 1\) and \(n_2 = \infty\):

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

The shortest wavelength (from energy levels \(\infty\) to 1) is \(\lambda\).

**Step 2: Calculating longest wavelength in the Balmer series for He⁺.**

For He⁺ (Z=2), in the Balmer series where \(n_1 = 2\) and \(n_2 = 3\):

\[ \frac{1}{\lambda_{\text{Balmer}}} = R \times 4 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

\[ \frac{1}{\lambda_{\text{Balmer}}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} \]

So, the wavelength for the longest transition in the Balmer series is:

\[ \lambda_{\text{Balmer}} = \frac{9}{5R} \]

**Relating it to the shortest wavelength of Hydrogen Lyman series.**

The shortest wavelength of the Lyman series (for hydrogen) is:

\[ \frac{1}{\lambda} = R \Rightarrow \lambda = \frac{1}{R} \]

We can relate \(\lambda_{\text{Balmer}}\) for He⁺ with \(\lambda\) of hydrogen:

\[ \lambda_{\text{Balmer}} = \frac{9\lambda}{5} \]

Thus, the correct option is \(\frac{9\lambda}{5}\), which matches the option "9λ/5".

**Conclusion:**

The longest wavelength in the Balmer series of He⁺ is \(\frac{9\lambda}{5}\), making the correct option "9λ/5".