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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2020 · 8 January · Morning Shift · Physics · Dual Nature of Matter

When photon of energy 4.0eV strikes the surface of a metal A, the...

When photon of energy 4.0eV4.0eV strikes the surface of a metal A,A, the ejected photoelectrons have maximum kinetic energy TAeV{T}_{A}eV and de-Broglie wavelength λA.{\lambda}_{A} . The maximum kinetic energy of photoelectrons liberated from another metal BB by photon of energy 4.50eV4.50eV is TB=(TA−1.5)eV.{T}_{B} = \left({T}_{A} - 1.5\right)eV . If the de-Broglie wavelength of these photoelectrons λB=2λA,{\lambda}_{B} = 2{\lambda}_{A}, then the work function of metal BB is:

Options

  1. A4eV4eV
  2. B2eV2eV
  3. C1.5eV1.5eV
  4. D3eV3eV

Correct answer

A. 4eV4eV

Step-by-step solution

Relation between de-Broglie wavelength and K.E. is
λ=h2(KE)me⇒λ∝1KE\lambda = \dfrac{h}{\sqrt{2 ( KE ) {m}_{e}}} ⇒ \lambda ∝ \dfrac{1}{\sqrt{KE}}
λAλB=KEBKEA\dfrac{{\lambda}_{A}}{{\lambda}_{B}} = \dfrac{\sqrt{K{E}_{B}}}{\sqrt{K{E}_{A}}}
⇒12=TA−1.5TA⇒TA=2eV⇒ \dfrac{1}{2} = \sqrt{\dfrac{{T}_{A} - 1.5}{{T}_{A}}} ⇒ {T}_{A} = 2eV
∴KEB=2−1.5=0.5eV∴ K{E}_{B} = 2 - 1.5 = 0.5eV
ϕB=4.5−0.5=4eV{ϕ}_{B} = 4.5 - 0.5 = 4eV
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