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JEE Main 2020 (8 Jan, Shift 1) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2020 (8 Jan, Shift 1)

Q126JEE Main 2020 · 8 JAN · SHIFT 1Difficult

Photoelectric Effect: Einstein's Equation

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Broglie wavelength λA\lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB\lambda_B = 2λA\lambda_A, then the work function of metal B is:

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Answer: (b)

  1. λ=h2mT⇒λ∝1T\lambda = \frac{h}{\sqrt{2mT}} \Rightarrow \lambda \propto \frac{1}{\sqrt{T}}; λB=2λA⇒TB=TA4\lambda_B = 2\lambda_A \Rightarrow T_B = \frac{T_A}{4}.
  2. TA−1.5=TA4⇒TA=2T_A - 1.5 = \frac{T_A}{4} \Rightarrow T_A = 2 eV, TB=0.5T_B = 0.5 eV
  3. ϕB=4.5−0.5=4\phi_B = 4.5 - 0.5 = 4 eV
  4. So the answer is (b) 4 eV.
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