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Physics Question 41 – JEE-MAIN 2026

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Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α×10−19 J. The value of α is _______. (h=6.62×10−34 J s, e=1.6×10−19 C and c=3×108 m/s)

The photoelectric effect describes the emission of electrons when light shines on a material.

Step 1: Identify the Governing Principle and Formula✦ Active

The problem involves the photoelectric effect. According to Einstein's photoelectric equation, the energy of the incident photon (E) is used to overcome the work function (Φ) of the metal and provide maximum kinetic energy (Kmax) to the emitted photoelectrons. The relevant formulas are:

E=Φ+Kmax E=hcλ Kmax=eVs

From these, we can express the work function as Φ=hcλ−eVs.

Step 2: Calculate Incident Photon Energy and Maximum Kinetic Energy○ Expand

Given values: λ=331 nm=331×10−9 m, Vs=0.2 V, h=6.62×10−34 J s, e=1.6×10−19 C, c=3×108 m/s.

Calculate the photon energy (E):

E=(6.62×10−34 J s)×(3×108 m/s)331×10−9 m=19.86×10−26331×10−9 J=0.06×10−17 J=6×10−19 J

Calculate the maximum kinetic energy (Kmax):

Kmax=(1.6×10−19 C)×(0.2 V)=0.32×10−19 J
💡 Teacher's Secret Hint

Ensure correct unit conversions, especially for wavelength from nm to m.

Step 3: Calculate the Work Function○ Expand

Now, substitute the calculated values into the work function formula:

Φ=E−Kmax=(6×10−19 J)−(0.32×10−19 J) Φ=(6−0.32)×10−19 J=5.68×10−19 J

The problem states the work function is α×10−19 J. Comparing this with our result, we find α=5.68.

💡 Teacher's Secret Hint

Pay attention to the required format of the final answer, specifically the value of α.

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