JEE Physics 2023 Question with Solution

Given: $A$, $B$ and $C$ are the first, second and third excited states of hydrogen, so they correspond to $n = 2$, $n = 3$ and $n = 4$ respectively.

Find: The value of $n$ from $\dfrac{\lambda_1}{\lambda_2} = \dfrac{7}{4n}$.

From the diagram, $\lambda_1$ corresponds to the transition $3 \rightarrow 2$ and $\lambda_2$ corresponds to the transition $4 \rightarrow 3$.

Using the Rydberg formula:

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

For $\lambda_1$:

$$\frac{1}{\lambda_1} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{5}{36}\right)$$

So,

$$\lambda_1 = \frac{36}{5R}$$

For $\lambda_2$:

$$\frac{1}{\lambda_2} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{7}{144}\right)$$

So,

$$\lambda_2 = \frac{144}{7R}$$

Now,

$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{36}{5R}}{\frac{144}{7R}} = \frac{36 \times 7}{144 \times 5} = \frac{7}{20}$$

Given,

$$\frac{\lambda_1}{\lambda_2} = \frac{7}{4n}$$

Therefore,

$$\frac{7}{4n} = \frac{7}{20}$$

So,

$$4n = 20 \Rightarrow n = 5$$

Therefore, the value of $n$ is $5$.

Given: $\lambda_1$ is for transition $3 \rightarrow 2$ and $\lambda_2$ is for transition $4 \rightarrow 3$.

Find: The value of $n$ in $\dfrac{\lambda_1}{\lambda_2} = \dfrac{7}{4n}$.

Instead of calculating wavelengths first, compare the reciprocals directly:

$$\frac{1}{\lambda_1} \propto \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36}$$ $$\frac{1}{\lambda_2} \propto \left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7}{144}$$

Hence,

$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\lambda_2}}{\frac{1}{\lambda_1}} = \frac{\frac{7}{144}}{\frac{5}{36}} = \frac{7}{20}$$

Comparing with

$$\frac{7}{4n}$$

we get

$$4n = 20$$

Therefore, $n = 5$.

This shortcut works because wavelength is inversely proportional to the transition energy term in the Rydberg relation.