JEE Main 2025 — Structure Of Atom Question with Solution

To determine the correct statements regarding the threshold frequency of cesium, first note that the threshold frequency given is $5.16\times 10^{14}\text{Hz}$. Using this frequency, we can calculate the corresponding wavelength, $\lambda$, using the equation:

$\lambda =\frac{c}{v}=\frac{3\times 10^8\text{m/s}}{5.16\times 10^{14}\text{Hz}}$

This calculation gives $\lambda \approx 581.39\text{nm}$.

Key points:

The wavelength calculated is near the yellow light region. Therefore, yellow light has enough energy to cause a photoelectric effect.

When the intensity of light (brightness) decreases, the photocurrent will also decrease, as fewer photons are available to eject electrons.

Red light, with its larger wavelength and lower frequency, does not have enough energy to cause the photoelectric effect in cesium.

Blue light has a higher frequency than yellow light ($v_{\text{blue}}>v_{\text{yellow}}$), meaning it can produce a photoelectric current.

White light includes all visible frequencies, including those above the threshold, so it will also emit a current.

Thus, the correct statements involve using yellow, blue, and white light, without producing an effect with red light:

A: Yellow light produces a current.

B: Dimming yellow light reduces the current.

D: Blue light produces a current.

E: White light produces a current.

Therefore, the correct statements are A, B, D, and E.