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JEE Main 2021 (25 Feb, Shift 2) · Photoelectric Effect: Einstein's Equation

Photoelectric Effect: Einstein's Equation question from JEE Main 2021 (25 Feb, Shift 2)

Q110JEE Main 2021 · 25 FEB · SHIFT 2Medium

Photoelectric Effect: Einstein's Equation

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:

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Answer: (d)

  1. Use E(eV)=1240λ(nm)E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}
  2. E1=1240491=2.525 eV⇒ϕ=2.525−0.710=1.815 eVE_1 = \frac{1240}{491} = 2.525\text{ eV} \Rightarrow \phi = 2.525 - 0.710 = 1.815\text{ eV}
  3. New photon energy =1.815+1.43=3.245 eV= 1.815 + 1.43 = 3.245\text{ eV}
  4. λ=12403.245=382 nm\lambda = \frac{1240}{3.245} = 382\text{ nm}
  5. So the answer is (d) 382 nm.
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