JEE Physics 2024 Question with Solution

Given: Energy of incident UV light is $4.13 \, \text{eV}$ and work function of the metal is $3.13 \, \text{eV}$.

Find: The maximum kinetic energy of the ejected photoelectrons.

Using Einstein's photoelectric equation:

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

Substituting the given values:

$$K_{\text{max}} = 4.13 \, \text{eV} - 3.13 \, \text{eV}$$ $$K_{\text{max}} = 1 \, \text{eV}$$

Therefore, the maximum kinetic energy of the ejected photoelectrons is $1 \, \text{eV}$. The correct option is B.

Given: Incident photon energy is $4.13 \, \text{eV}$ and work function is $3.13 \, \text{eV}$.

Find: Maximum kinetic energy.

In the photoelectric effect, the emitted electron gets the leftover energy after overcoming the work function. So subtract the work function from the incident photon energy:

$$K.E. = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV}$$

Therefore, the correct option is B.