JEE Physics 2023 Question with Solution

Given: Radiation of wavelength $\lambda_0$ is emitted for the transition from second excited state to first excited state, that is $n=3 \to n=2$. For the second transition, the electron goes from third excited state to second orbit, that is $n=4 \to n=2$.

Find: The value of $x$ in $\lambda = \frac{20}{x}\lambda_0$.

Using the Rydberg relation for hydrogen:

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

For the first transition $n=3 \to n=2$,

$$\frac{1}{\lambda_0} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$$ $$\frac{1}{\lambda_0} = R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{5}{36}\right)$$

For the second transition $n=4 \to n=2$,

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

Now take the ratio:

$$\frac{\lambda_0}{\lambda} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{5\times 16}{36\times 3} = \frac{20}{27}$$

Hence,

$$\lambda = \frac{20}{27}\lambda_0$$

Comparing with $\lambda = \frac{20}{x}\lambda_0$, we get

$$x = 27$$

Therefore, the value of $x$ is $27$.

Given: Two hydrogen atom transitions, $3 \to 2$ and $4 \to 2$.

Find: The number $x$ in $\frac{20}{x}\lambda_0$.

For hydrogen, wavelength is inversely proportional to the energy gap, so

$$\frac{1}{\lambda_0} \propto \left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5}{36}$$

and

$$\frac{1}{\lambda} \propto \left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3}{16}$$

Therefore,

$$\frac{\lambda_0}{\lambda} = \frac{\frac{5}{36}}{\frac{3}{16}} = \frac{20}{27}$$

So,

$$\lambda = \frac{20}{27}\lambda_0$$

Thus, the correct numerical value is $27$.