Question

In hydrogen spectrum, the frequency of the spectral line corresponding to electron transition $n_2 = 3$ to $n_1 = 2$ is $x$ Hz. What is the frequency (in Hz) of the spectral line corresponding to electron transition $n_2 = 4$ to $n_1 = 3$ of $\text{He}^+$ spectrum?

• A. $\frac{5x}{7}$
• B. $\frac{7x}{5}$
• C. $\frac{20x}{7}$
• D. $\frac{7x}{20}$

Hint:

Always write out the ratio of frequencies $\frac{\nu_1}{\nu_2} = \frac{Z_1^2}{Z_2^2} \times \frac{(1/n_{1a}^2 - 1/n_{2a}^2)}{(1/n_{1b}^2 - 1/n_{2b}^2)}$ to directly eliminate constants.
This reduces algebraic calculations and avoids units errors.

Answer 1

The Correct Option is B

Step 1: Set up energies instead of frequencies.
For a hydrogen-like ion the energy of a level is $E_n = -13.6\, Z^2/n^2$ eV, so the photon energy for a transition is $\Delta E = 13.6 Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$, and since $\Delta E = h\nu$, the frequency scales the same way with $Z^2$ and the level terms.
Step 2: Write the hydrogen case.
For H, $Z=1$, $n_1=2$, $n_2=3$: $\dfrac{1}{4}-\dfrac{1}{9}=\dfrac{5}{36}$, so $x \propto 1^2 \times \dfrac{5}{36}$.
Step 3: Write the He+ case.
For $\text{He}^+$, $Z=2$, $n_1=3$, $n_2=4$: $\dfrac{1}{9}-\dfrac{1}{16}=\dfrac{7}{144}$, so the new frequency is proportional to $2^2 \times \dfrac{7}{144} = \dfrac{7}{36}$.
Step 4: Take the ratio.
Dividing the two proportional expressions cancels the common constant $cR$: $\dfrac{\nu}{x} = \dfrac{7/36}{5/36} = \dfrac{7}{5}$.
\[ \boxed{\nu = \dfrac{7x}{5}} \]