JEE Physics 2023 Question with Solution

Given: The wavelength of incident radiation is $350 \, \text{nm}$. The work functions are $\phi_A = 4.8 \, \text{eV}$ and $\phi_B = 2.2 \, \text{eV}$.

Find: Which metal emits photo-electrons.

For photoelectric emission to occur, the photon energy must be greater than the work function of the material.

The photon energy is given by

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

Substituting $\lambda = 350 \times 10^{-9} \, \text{m}$,

$$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{350 \times 10^{-9}} = 5.69 \, \text{eV}$$

Comparing this with the work functions:

• For metal A, $5.69 \, \text{eV} > 4.8 \, \text{eV}$
• For metal B, $5.69 \, \text{eV} > 2.2 \, \text{eV}$

So both metals should emit photo-electrons. However, the provided the solution concludes: The Correct Option is D and states that metal A will not emit photo-electrons, which is inconsistent with the numerical comparison shown.

Based on the source solution authority, the correct option is D.

Given: Incident wavelength $350 \, \text{nm}$.

Find: Whether the listed answer matches the shown working.

Using the photoelectric condition,

$$E_{photon} \ge \phi$$

From the extracted working,

$$E_{photon} = 5.69 \, \text{eV}$$

Now compare with each work function:

$$\phi_A = 4.8 \, \text{eV}$$ $$\phi_B = 2.2 \, \text{eV}$$

Since $5.69 > 4.8$ and $5.69 > 2.2$, both metals satisfy the emission condition. Therefore, the numerical working supports option B, not option D.

The solution's nevertheless explicitly marks D as correct. Hence the extracted answer is kept as D, with this discrepancy noted.