JEE Physics 2024 Question with Solution

Given: Work function of the substance is $\phi = 3.0 \, \text{eV}$.

Find: The longest wavelength $\lambda_{\max}$ that can emit photoelectrons.

For photoelectric emission at threshold,

$$\phi = \frac{hc}{\lambda_{\max}}$$

So,

$$\lambda_{\max} = \frac{hc}{\phi}$$

Using $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$, we get

$$\phi = 3.0 \times 1.602 \times 10^{-19} \, \text{J} = 4.806 \times 10^{-19} \, \text{J}$$

Now substitute

$$\lambda_{\max} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{4.806 \times 10^{-19}}$$ $$\lambda_{\max} \approx 4.14 \times 10^{-7} \, \text{m} = 414 \, \text{nm}$$

Therefore, the longest wavelength is $414 \, \text{nm}$. The correct option is B.

The solution also shows a contradictory label "The Correct Option is C", but the working and final numerical conclusion clearly give $414 \, \text{nm}$, which matches option B.

Given: $\phi = 3.0 \, \text{eV}$

Find: Threshold wavelength $\lambda_{\max}$.

Use the direct relation in electron-volt units:

$$\lambda_{\max} = \frac{1240 \, \text{nm} \cdot \text{eV}}{3.0 \, \text{eV}}$$ $$\lambda_{\max} = 413.33 \, \text{nm} \approx 414 \, \text{nm}$$

This shortcut works because $hc$ is commonly used as $1240 \, \text{nm} \cdot \text{eV}$, avoiding conversion to joules.

Therefore, the correct option is B.