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Dual Nature of Radiation question

2024 · 9 Apr · Shift 2 · Q69
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Dual Nature of Radiation question

2024 · 9 Apr · Shift 2 · Q69

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
UV light of 4.13 eV4.13 \mathrm{~eV}4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV3.13 \mathrm{~eV}3.13 eV. The maximum kinetic energy of ejected photoelectrons will be:
  1. A
    4.13 eV
  2. B
    1 eV
  3. C
    7.26 eV
  4. D
    3.13 eV
View written solutionFree

Correct answer: B

  1. Use Einstein’s photoelectric equation

    The maximum kinetic energy of emitted photoelectrons is Kmax⁡=hν−ϕK_{\max} = h\nu - \phiKmax​=hν−ϕ where:

    • hνh\nuhν = energy of incident photon
    • ϕ\phiϕ = work function of the metal

  2. Substitute the given values

    Given: hν=4.13 eVh\nu = 4.13\,\text{eV}hν=4.13eV ϕ=3.13 eV\phi = 3.13\,\text{eV}ϕ=3.13eV

    Therefore, Kmax⁡=4.13−3.13=1.00 eVK_{\max} = 4.13 - 3.13 = 1.00\,\text{eV}Kmax​=4.13−3.13=1.00eV

  3. Match with the options

    Kmax⁡=1 eVK_{\max} = 1\,\text{eV}Kmax​=1eV

    So the correct option is B.

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