Verified Solution • 2025–26 Session moderate

QJEEM20Ph403a: JEE Main 2020 Physics - Class 12 JEEM20Ph403a Solution: When photon of energy 4.0 eV strikes the surface of a metal A,…

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength λA. 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 = 2λA, then the work function of metal B is :

Multiple Choice Options

  • A 1.5eV
  • B 4eV
  • C 2eV
  • D 3eV

Step-by-Step Verified Solution

We know, de-Broglie wavelength

λ = hp=h2mKe

∴ λ∝1Ke

So λAλB=TBTA

⇒ (12)2 = TA−1.5TA

On solving TA = 2 eV

∴ TB = TA - 1.5 = 0.5 eV

Also TB = 4.5 - ϕB

⇒ 0.5 = 4.5 - ϕB

⇒ ϕB = 4 eV

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