Question

The ratio of wavelengths for transition of electrons from 2nd orbit to $1^{st}$ orbit of Helium $(He^{+})$ and Lithium $(Li^{++})$ is (Atomic number of Helium = 2, Atomic number of Lithium = 3)

• A. 9:4
• B. 9:36
• C. 4:9
• D. 2:3

Hint:

Logic Tip: Wavelength is inversely proportional to energy ($E = hc/\lambda$). A higher atomic number $Z$ means a stronger pull on the electrons, resulting in a much larger energy gap between orbits ($E \propto Z^2$). Since Lithium has a larger $Z$, it emits a much higher energy photon, meaning a much shorter wavelength. Thus, the $He$ to $Li$ wavelength ratio must be $> 1$, eliminating options C and D immediately.

Answer 1

The Correct Option is A

Concept:
The wavelength ($\lambda$) of a photon emitted during an electronic transition in a hydrogen-like atom or ion is given by the Rydberg formula: $$\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ where $R$ is the Rydberg constant, $Z$ is the atomic number, $n_1$ is the lower energy level, and $n_2$ is the higher energy level.
Step 1: Set up the ratio for the two ions.
Since the transition is the same for both ions (from $n_2 = 2$ to $n_1 = 1$), the term $R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ is constant for both. Therefore, the reciprocal of the wavelength is directly proportional to the square of the atomic number ($Z^2$): $$\frac{1}{\lambda} \propto Z^2 \implies \lambda \propto \frac{1}{Z^2}$$
Step 2: Calculate the ratio of the wavelengths.
For Helium ($He^+$), $Z_{He} = 2$. For Lithium ($Li^{++}$), $Z_{Li} = 3$. The ratio of their wavelengths is: $$\frac{\lambda_{He{\lambda_{Li = \frac{Z_{Li}^2}{Z_{He}^2}$$ Substitute the atomic numbers: $$\frac{\lambda_{He{\lambda_{Li = \frac{3^2}{2^2} = \frac{9}{4}$$ Thus, the ratio is 9:4.