JEE Physics 2024 Question with Solution

Given: Rydberg’s constant is $R$ and the longest wavelength in the Paschen series is to be written as $\frac{\alpha}{7R}$.

Find: The value of $\alpha$.

The Paschen series corresponds to transitions ending at $n = 3$.

For the longest wavelength, the energy difference must be minimum, so the transition is from $n_2 = 4$ to $n_1 = 3$.

Using the Rydberg formula:

$$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

Substituting $n_1 = 3$ and $n_2 = 4$:

$$\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right)$$ $$\frac{1}{\lambda} = R\left(\frac{1}{9} - \frac{1}{16}\right)$$ $$\frac{1}{\lambda} = R\left(\frac{16-9}{144}\right) = \frac{7R}{144}$$

Hence,

$$\lambda = \frac{144}{7R}$$

Comparing with $\frac{\alpha}{7R}$, we get $\alpha = 144$.

Therefore, the required value is $144$.

Given: The radiation belongs to the Paschen series.

Find: The value of $\alpha$ in $\lambda = \frac{\alpha}{7R}$.

In any spectral series, the longest wavelength corresponds to the smallest allowed transition in that series.

For Paschen series, electrons fall to $n = 3$, so the first line is $4 \to 3$.

Directly write:

$$\frac{1}{\lambda} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$$

Therefore,

$$\lambda = \frac{144}{7R}$$

So, the correct value of $\alpha$ is $144$.