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Physics Question 43 – JEE-MAIN 2025

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The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

For photoemission to occur, the energy of the incident photon must be greater than or equal to the work function of the metal.

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Ninja Strategy•Energy-Wavelength Relationship

Photoelectric emission requires photon energy greater than the work function. Since photon energy increases with decreasing wavelength (from Red to Violet), the highest energy option (Blue) is the most likely candidate to cause emission.

Step 1: Determine the threshold energy and wavelength✦ Active

The work function of the metal is ϕ=3 eV. For photoelectron emission, the energy of the incident photon E must satisfy E≥ϕ. The threshold wavelength λth is the maximum wavelength of light that can cause emission, corresponding to the minimum energy ϕ.

λth=hcϕ

Using the relation E=hcλ, where hc≈1240 eV nm, we find the threshold wavelength:

λth=1240 eV nm3 eV≈413.3 nm
Step 2: Compare threshold wavelength with visible light spectrum○ Expand

For emission to occur, the incident light must have a wavelength λ≤λth, i.e., λ≤413.3 nm. The visible light spectrum ranges approximately from 380 nm (violet) to 750 nm (red). The order of colors by increasing energy (decreasing wavelength) is Red, Yellow, Green, Blue, Violet. Typical wavelength ranges are:

Red: 620−750 nm

Yellow: 570−590 nm

Green: 495−570 nm

Blue: 450−495 nm

Violet: 380−450 nm

Step 3: Identify the suitable color from options○ Expand

The required wavelength λ≤413.3 nm falls within the violet part of the spectrum. Among the given options (Yellow, Green, Red, Blue), Blue light has the shortest wavelength and highest energy. While 413.3 nm is strictly in the violet range, "Blue" is the closest and highest energy option provided, often used to encompass the shorter wavelength end of the visible spectrum when "Violet" is not an option. Therefore, Blue light is the most appropriate choice.

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