JEE Physics 2025 Question with Solution

Given: Initial maximum kinetic energy is $2 \, \text{eV}$ and the work function is $1 \, \text{eV}$.

Find: The new maximum kinetic energy when the wavelength becomes $\frac{\lambda}{2}$.

Use Einstein's photoelectric equation:

$$K_{\max} = h\nu - \phi$$

For wavelength $\lambda$:

$$h\nu = K_{\max} + \phi = 2 + 1 = 3 \, \text{eV}$$

If the wavelength is halved, photon energy doubles:

$$h\nu' = \frac{hc}{\lambda/2} = 2\frac{hc}{\lambda} = 2 \times 3 = 6 \, \text{eV}$$

Now apply the photoelectric equation again:

$$K'_{\max} = h\nu' - \phi = 6 - 1 = 5 \, \text{eV}$$

Therefore, the maximum kinetic energy is $5 \, \text{eV}$ and the correct option is B.

Given: For the first light source, the emitted electrons have maximum kinetic energy $2 \, \text{eV}$ and the work function is $1 \, \text{eV}$.

Find: The kinetic energy for wavelength $\frac{\lambda}{2}$.

Since photon energy is inversely proportional to wavelength, halving the wavelength doubles the photon energy. The initial photon energy is:

$$E = K_{\max} + \phi = 2 + 1 = 3 \, \text{eV}$$

So the new photon energy is:

$$E' = 2E = 6 \, \text{eV}$$

Subtract the same work function:

$$K'_{\max} = 6 - 1 = 5 \, \text{eV}$$

Therefore, the correct option is B.