JEE Physics 2025 Question with Solution

Given: The wavelength for transition $A \to C$ is $\lambda_{AC} = 2000 \, \text{Å}$ and for transition $B \to C$ is $\lambda_{BC} = 6000 \, \text{Å}$.

Find: The wavelength $\lambda_{AB}$ for transition $A \to B$.

For an emitted photon,

$$E = \frac{hc}{\lambda}$$

So the transition energy is inversely proportional to wavelength.

From the energy level relation,

$$E_{AB} = E_{AC} - E_{BC}$$

Therefore,

$$\frac{hc}{\lambda_{AB}} = \frac{hc}{\lambda_{AC}} - \frac{hc}{\lambda_{BC}}$$

Substituting the given wavelengths,

$$\frac{hc}{\lambda_{AB}} = \frac{hc}{2000 \, \text{Å}} - \frac{hc}{6000 \, \text{Å}}$$

Cancelling $hc$,

$$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000}$$

Taking the common denominator,

$$\frac{1}{\lambda_{AB}} = \frac{3 - 1}{6000} = \frac{2}{6000} = \frac{1}{3000}$$

Hence,

$$\lambda_{AB} = 3000 \, \text{Å}$$

Therefore, the wavelength emitted in the transition from $A$ to $B$ is $3000 \, \text{Å}$. The correct option is A.

Given: $\lambda_{AC} = 2000 \, \text{Å}$ and $\lambda_{BC} = 6000 \, \text{Å}$.

Find: $\lambda_{AB}$.

Since photon energy is proportional to $\frac{1}{\lambda}$, the transition energies satisfy

$$E_{AC} = E_{AB} + E_{BC}$$

So directly,

$$\frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}}$$

Substitute the values:

$$\frac{1}{2000} = \frac{1}{\lambda_{AB}} + \frac{1}{6000}$$

Thus,

$$\frac{1}{\lambda_{AB}} = \frac{1}{2000} - \frac{1}{6000} = \frac{1}{3000}$$

Hence,

$$\lambda_{AB} = 3000 \, \text{Å}$$

This works because level energy differences add directly, while wavelength must be handled through its reciprocal. Therefore, the correct option is A.