Question

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the photons emitted when an electron in the $n^{\text{th}}$ orbit of a hydrogen atom falls to the first excited state and the ground state respectively, then the value of $n$ is

• A. $\sqrt{\dfrac{2(\lambda_2-\lambda_1)}{2\lambda_2-\lambda_1}}$
• B. $\dfrac{2\lambda_2-\lambda_1}{2(\lambda_2-\lambda_1)}$
• C. $\sqrt{\dfrac{4\lambda_2-\lambda_1}{4(\lambda_2-\lambda_1)}}$
• D. $\sqrt{\dfrac{4(\lambda_2-\lambda_1)}{4\lambda_2-\lambda_1}}$

Hint:

Always write separate Rydberg equations for each transition and eliminate the Rydberg constant to relate wavelengths.

Answer 1

The Correct Option is D

Step 1: Use the Rydberg formula for hydrogen: \[ \frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) \]
Step 2: For transition from $n$ to first excited state ($n_f=2$): \[ \frac{1}{\lambda_1} =R\left(\frac{1}{2^2}-\frac{1}{n^2}\right) =R\left(\frac{1}{4}-\frac{1}{n^2}\right) \quad \cdots (1) \]
Step 3: For transition from $n$ to ground state ($n_f=1$): \[ \frac{1}{\lambda_2} =R\left(1-\frac{1}{n^2}\right) \quad \cdots (2) \]
Step 4: Subtract equation (1) from (2): \[ \frac{1}{\lambda_2}-\frac{1}{\lambda_1} =R\left(1-\frac{1}{4}\right) =\frac{3R}{4} \]
Step 5: Eliminate $R$ using equation (2): \[ R=\frac{1}{\lambda_2\left(1-\frac{1}{n^2}\right)} \] Substitute into Step 4 and simplify to obtain: \[ n^2=\frac{4(\lambda_2-\lambda_1)}{4\lambda_2-\lambda_1} \]
Step 6: Hence, \[ n=\sqrt{\frac{4(\lambda_2-\lambda_1)}{4\lambda_2-\lambda_1}} \]