Question

Light of wavelength \(\frac{36}{5R}\,\text{m}\) is emitted by a hydrogen atom during the transition of electrons from the state

• A. \(n=3\) to \(n=2\)
• B. \(n=4\) to \(n=1\)
• C. \(n=4\) to \(n=2\)
• D. \(n=4\) to \(n=3\)
• E. \(n=3\) to \(n=1\)

Hint:

In hydrogen spectrum questions, first convert the given wavelength into \(1/\lambda\), then compare it directly with the Rydberg expression \(\frac{1}{n_1^2}-\frac{1}{n_2^2}\).

Answer 1

The Correct Option is A

Step 1: Recall the Rydberg formula for hydrogen spectrum.
For a transition from higher level \(n_2\) to lower level \(n_1\), the emitted wavelength satisfies:
\[ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] where \(R\) is the Rydberg constant.

Step 2: Use the wavelength given in the question.
The wavelength is:
\[ \lambda=\frac{36}{5R} \] Therefore, \[ \frac{1}{\lambda}=\frac{5R}{36} \]

Step 3: Substitute into the Rydberg formula.
So we must have: \[ R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{5R}{36} \] Cancelling \(R\) from both sides: \[ \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{5}{36} \]

Step 4: Test option \(n=3\) to \(n=2\).
Here \(n_2=3\) and \(n_1=2\). Then: \[ \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{2^2}-\frac{1}{3^2} \] \[ =\frac{1}{4}-\frac{1}{9} \] \[ =\frac{9-4}{36}=\frac{5}{36} \] This exactly matches the required value.

Step 5: Verify that this corresponds to emission.
Since the electron moves from \(n=3\) to \(n=2\), it goes from higher to lower energy level, so light is emitted. Hence the transition is physically correct.

Step 6: Compare with the remaining options.
Since the first option gives the exact factor \(\frac{5}{36}\), there is no need to test further. Therefore, the correct transition is uniquely identified.

Step 7: State the final answer.
Thus, the emitted wavelength \(\frac{36}{5R}\) corresponds to the transition:
\[ \boxed{n=3 \text{ to } n=2} \] which matches option \((1)\).