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Dual Nature of Radiation and Matter · Class 12 · CBSE · Science Physics

When the wavelength of incident light on a photosensitive surface is changed from lambda1 to lambda2, the stopping potential for photoelectrons chang…

Practice question from Dual Nature of Radiation and Matter (Class 12 · CBSE · Science Physics), with the correct answer and a step-by-step solution.

hard · MCQ · Q-915ECC

When the wavelength of incident light on a photosensitive surface is changed from λ1\lambda_1 to λ2\lambda_2, the stopping potential for photoelectrons changes from V1V_1 to V2V_2. The value of the specific charge (e/me/m) of the electron cannot be determined directly, but the ratio of threshold frequency to work function is independent. Which of the following expressions correctly represents the Planck's constant hh?

  1. A.e(V1−V2)c(λ2−λ1)\frac{e(V_1 - V_2)}{c(\lambda_2 - \lambda_1)}
  2. B.e(V1−V2)(λ1λ2)c(λ2−λ1)\frac{e(V_1 - V_2)(\lambda_1 \lambda_2)}{c(\lambda_2 - \lambda_1)}Correct answer
  3. C.e(V1−V2)(λ2−λ1)c(λ1λ2)\frac{e(V_1 - V_2)(\lambda_2 - \lambda_1)}{c(\lambda_1 \lambda_2)}
  4. D.hce(λ2−λ1)\frac{hc}{e(\lambda_2 - \lambda_1)}

Solution

Using Einstein's photoelectric equation: eV1=hcλ1−ϕeV_1 = \frac{hc}{\lambda_1} - \phi and eV2=hcλ2−ϕeV_2 = \frac{hc}{\lambda_2} - \phi. Subtracting the two: e(V1−V2)=hc(1λ1−1λ2)=hc(λ2−λ1λ1λ2)e(V_1 - V_2) = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) = hc \left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right). Solving for hh gives h=e(V1−V2)(λ1λ2)c(λ2−λ1)h = \frac{e(V_1 - V_2)(\lambda_1 \lambda_2)}{c(\lambda_2 - \lambda_1)}.

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