Question

The frequency of the light emitted when an electron comes down from \(n=4\) to \(n=2\) level in hydrogen atom is \(\dfrac{3}{7}\) times of the following transition of the Li atom:

• A. \(4\) to \(3\)
• B. \(4\) to \(1\)
• C. \(3\) to \(2\)
• D. \(5\) to \(3\)

Hint:

For hydrogen-like atoms, frequency is proportional to \[ Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right). \] Always include the \(Z^2\) factor for ions like lithium.

Answer 1

The Correct Option is A

Step 1: Write the Rydberg frequency formula for hydrogen-like atoms.
For a hydrogen-like ion with atomic number $Z$:
\[ \nu = RcZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] where $R$ is the Rydberg constant, $c$ is speed of light, and the transition is from level $n_2$ to $n_1$ ($n_1 < n_2$).

Step 2: Calculate frequency for H atom transition $4 \to 2$.
For H, $Z = 1$, $n_1 = 2$, $n_2 = 4$:
\[ \nu_H = Rc(1)^2\left(\frac{1}{4} - \frac{1}{16}\right) = Rc\left(\frac{4-1}{16}\right) = \frac{3Rc}{16} \]

Step 3: Find the required Li frequency.
Given $\nu_H = \frac{3}{7}\nu_{\text{Li}}$:
\[ \nu_{\text{Li}} = \frac{7}{3}\nu_H = \frac{7}{3} \times \frac{3Rc}{16} = \frac{7Rc}{16} \]

Step 4: Set up the equation for Li (Z=3) transition.
For Li (as a hydrogen-like ion), $Z = 3$:
\[ \nu_{\text{Li}} = Rc \times 9 \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = \frac{7Rc}{16} \] \[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{7}{144} \]

Step 5: Test the transition $4 \to 3$ for Li.
\[ \frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \checkmark \] This matches exactly.

Step 6: State the answer.
\[ \boxed{4 \text{ to } 3} \]