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JEE Main 2016 (9 Apr, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2016 (9 Apr, Shift 1)

Q144JEE Main 2016 · 9 APR · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation

When photons of wavelength λ1\lambda_1 are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ2\lambda_2 are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ3\lambda_3 is used then find the stopping potential for this case:

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Answer: (c)

  1. eV=hcλ1−ϕeV = \frac{hc}{\lambda_1} - \phi and 3eV=hcλ2−ϕ3eV = \frac{hc}{\lambda_2} - \phi.
  2. Subtract: eV=hc2(1λ2−1λ1)eV = \frac{hc}{2}\left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right), so ϕ=hcλ1−eV=hc(32λ1−12λ2)\phi = \frac{hc}{\lambda_1} - eV = hc\left(\frac{3}{2\lambda_1} - \frac{1}{2\lambda_2}\right).
  3. V3=1e(hcλ3−ϕ)=hce[1λ3+12λ2−32λ1]V_3 = \frac{1}{e}\left(\frac{hc}{\lambda_3} - \phi\right) = \frac{hc}{e}\left[\frac{1}{\lambda_3} + \frac{1}{2\lambda_2} - \frac{3}{2\lambda_1}\right]
  4. So the answer is (c).
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