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

1. Energy Levels and Wavelengths: The energy of a transition in a hydrogen-like atom is given by:
$$E = -\frac{13.6 Z^2}{n^2},$$
where $Z$ is the atomic number and $n$ is the principal quantum number. The wavelength $\lambda$ is related to the energy difference between levels:
$$\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right),$$
where $R$ is the Rydberg constant, and $n_1$ and $n_2$ are the lower and higher energy levels, respectively.
2. Lyman Series (Hydrogen): The shortest wavelength in the Lyman series corresponds to a transition from $n_2 = \infty$ to $n_1 = 1$. For hydrogen ($Z = 1$):
$$\frac{1}{\lambda} = R(1)^2 \left( 1 - 0 \right) \implies \lambda = \frac{1}{R}.$$
3. Balmer Series (He$^+$): For He$^+$ ($Z = 2$), the longest wavelength in the Balmer series corresponds to a transition from $n_2 = 3$ to $n_1 = 2$. The wavelength is given by:
$$\frac{1}{\lambda_{\text{Balmer}}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right).$$
Simplify:
$$\frac{1}{\lambda_{\text{Balmer}}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{9} \right).$$
Thus:
$$\lambda_{\text{Balmer}} = \frac{9}{5}\frac{1}{\lambda}.$$
Final Answer: $\frac{9}{5}\lambda$.