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: Use the frequency formula for hydrogen-like atoms.
For a hydrogen-like atom, \[ \nu=RcZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] For hydrogen atom, \[ Z=1 \]

Step 2: Find frequency for hydrogen transition \(4\to 2\).
\[ \nu_H=Rc\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \] \[ \nu_H=Rc\left(\frac{1}{4}-\frac{1}{16}\right) \] \[ \nu_H=Rc\left(\frac{3}{16}\right) \]

Step 3: Use the given relation.
Given, \[ \nu_H=\frac{3}{7}\nu_{\text{Li}} \] So, \[ \nu_{\text{Li}}=\frac{7}{3}\nu_H \] \[ \nu_{\text{Li}}=\frac{7}{3}\times \frac{3Rc}{16} \] \[ \nu_{\text{Li}}=\frac{7Rc}{16} \]

Step 4: Check Li transition.
For Li atom, taking it as hydrogen-like lithium ion, \[ Z=3 \] So, \[ \nu_{\text{Li}}=9Rc\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] For transition \(4\to 3\), \[ \nu_{\text{Li}}=9Rc\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \] \[ \nu_{\text{Li}}=9Rc\left(\frac{1}{9}-\frac{1}{16}\right) \] \[ \nu_{\text{Li}}=9Rc\left(\frac{7}{144}\right) \] \[ \nu_{\text{Li}}=\frac{7Rc}{16} \] This matches the required frequency.

Step 5: Final conclusion.
Hence, the required transition is \[ \boxed{4\text{ to }3} \]