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Dual Nature of Radiation question

2022 · 24 Jun · Shift 2 · Q62
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Dual Nature of Radiation question

2022 · 24 Jun · Shift 2 · Q62

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :
  1. A
    1 : 1
  2. B
    2 : 1
  3. C
    4 : 1
  4. D
    1 : 4
View written solutionFree

Correct answer: B

  1. Use Einstein’s photoelectric equation

For photoelectric emission, Kmax⁡=hν−ϕK_{\max} = h\nu - \phiKmax​=hν−ϕ where:

  • Kmax⁡K_{\max}Kmax​ = maximum kinetic energy of emitted electron
  • hνh\nuhν = photon energy
  • ϕ\phiϕ = work function

Given work function: ϕ=0.6 eV\phi = 0.6\,\text{eV}ϕ=0.6eV

Photon energies are:

  • First light: 3.8 eV3.8\,\text{eV}3.8eV
  • Second light: 1.4 eV1.4\,\text{eV}1.4eV

  1. Find maximum kinetic energies

For the first frequency: K1=3.8−0.6=3.2 eVK_1 = 3.8 - 0.6 = 3.2\,\text{eV}K1​=3.8−0.6=3.2eV

For the second frequency: K2=1.4−0.6=0.8 eVK_2 = 1.4 - 0.6 = 0.8\,\text{eV}K2​=1.4−0.6=0.8eV


  1. Relate kinetic energy to speed

Since Kmax⁡=12mv2K_{\max} = \frac{1}{2}mv^2Kmax​=21​mv2 for electrons of the same mass, v∝Kv \propto \sqrt{K}v∝K​

Therefore, v1v2=K1K2=3.20.8=4=2\frac{v_1}{v_2} = \sqrt{\frac{K_1}{K_2}} = \sqrt{\frac{3.2}{0.8}} = \sqrt{4} = 2v2​v1​​=K2​K1​​​=0.83.2​​=4​=2

So the ratio of maximum speeds is: v1:v2=2:1v_1 : v_2 = 2 : 1v1​:v2​=2:1


  1. Check options
  • A: 1:11:11:1 ❌
  • B: 2:12:12:1 ✅
  • C: 4:14:14:1 ❌
  • D: 1:41:41:4 ❌

Hence, the correct option is B.

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