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JEE Physics 2026 Question with Solution

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V3.2 \, \text{V}. If a second light having wavelength twice of the first light is used, the stopping potential drops to 0.7 V0.7 \, \text{V}. The wavelength of the first light is _____ m\text{m}.

  • A

    2.2×10−82.2 \times 10^{-8}

  • B

    3.1×10−73.1 \times 10^{-7}

  • C

    2.5×10−72.5 \times 10^{-7}

  • D

    2.9×10−82.9 \times 10^{-8}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The stopping potentials are 3.2 V3.2 \, \text{V} and 0.7 V0.7 \, \text{V} for wavelengths λ\lambda and 2λ2\lambda respectively.

Find: The wavelength of the first light.

Use Einstein’s photoelectric equation:

eVs=hcλ−ϕeV_s = \frac{hc}{\lambda} - \phi

For the first light,

e(3.2)=hcλ−ϕe(3.2) = \frac{hc}{\lambda} - \phi

For the second light of wavelength 2λ2\lambda,

e(0.7)=hc2λ−ϕe(0.7) = \frac{hc}{2\lambda} - \phi

Subtract the second equation from the first:

e(3.2−0.7)=hcλ−hc2λe(3.2 - 0.7) = \frac{hc}{\lambda} - \frac{hc}{2\lambda}

So,

e(2.5)=hc2λe(2.5) = \frac{hc}{2\lambda}

Now solve for λ\lambda:

λ=hc2e(2.5)\lambda = \frac{hc}{2e(2.5)}

Substitute the constants:

λ=6.63×10−34×3×1082×1.6×10−19×2.5\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 1.6 \times 10^{-19} \times 2.5}

This gives

λ≈2.5×10−7 m\lambda \approx 2.5 \times 10^{-7} \, \text{m}

Therefore, the wavelength of the first light is 2.5×10−7 m2.5 \times 10^{-7} \, \text{m} and the correct option is C.

Common mistakes

  • Using intensity instead of frequency or wavelength to determine stopping potential. This is wrong because stopping potential depends on photon energy, not intensity. Use Einstein’s photoelectric equation with wavelength for each case.

  • Assuming that doubling the wavelength doubles the stopping potential change directly. This is wrong because photon energy is inversely proportional to wavelength. First write separate equations for λ\lambda and 2λ2\lambda, then subtract them.

  • Forgetting to eliminate the work function ϕ\phi by subtraction. This is wrong because ϕ\phi is unknown and common to both cases. Subtract the two photoelectric equations before substituting constants.

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