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Dual Nature of Radiation question

2020 · 8 Jan · Shift 1 · Q41
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Dual Nature of Radiation question

2020 · 8 Jan · Shift 1 · Q41

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
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\lambda _AλA​. 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 _BλB​ = 2 λA\lambda _AλA​, then the work function of metal B is :
  1. A
    1.5eV
  2. B
    4eV
  3. C
    2eV
  4. D
    3eV
View written solutionFree

Correct answer: B

  1. Use photoelectric equation for both metals

For metal AAA: TA=4.0−ϕAT_A = 4.0 - \phi_ATA​=4.0−ϕA​

For metal BBB: TB=4.50−ϕBT_B = 4.50 - \phi_BTB​=4.50−ϕB​

Given: TB=TA−1.5T_B = T_A - 1.5TB​=TA​−1.5

  1. Use de-Broglie wavelength relation

For a photoelectron, λ=h2mT  ⟹  λ∝1T\lambda = \frac{h}{\sqrt{2mT}} \implies \lambda \propto \frac{1}{\sqrt{T}}λ=2mT​h​⟹λ∝T​1​

Given: λB=2λA\lambda_B = 2\lambda_AλB​=2λA​

So, λBλA=2=TATB\frac{\lambda_B}{\lambda_A} = 2 = \sqrt{\frac{T_A}{T_B}}λA​λB​​=2=TB​TA​​​

Squaring both sides: 4=TATB4 = \frac{T_A}{T_B}4=TB​TA​​ TA=4TBT_A = 4T_BTA​=4TB​

  1. Combine with the kinetic energy condition

Given: TB=TA−1.5T_B = T_A - 1.5TB​=TA​−1.5

Substitute TA=4TBT_A = 4T_BTA​=4TB​: TB=4TB−1.5T_B = 4T_B - 1.5TB​=4TB​−1.5 3TB=1.53T_B = 1.53TB​=1.5 TB=0.5 eVT_B = 0.5\ \text{eV}TB​=0.5 eV

Then, TA=4TB=2.0 eVT_A = 4T_B = 2.0\ \text{eV}TA​=4TB​=2.0 eV

  1. Find work function of metal B

Using TB=4.50−ϕBT_B = 4.50 - \phi_BTB​=4.50−ϕB​

So, ϕB=4.50−0.50=4.0 eV\phi_B = 4.50 - 0.50 = 4.0\ \text{eV}ϕB​=4.50−0.50=4.0 eV

  1. Check options

Thus the work function of metal BBB is: 4 eV\boxed{4\ \text{eV}}4 eV​

So the correct option is B.

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