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Dual Nature of Radiation question

2022 · 27 Jun · Shift 1 · Q54
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  5. /2022 · 27 Jun · Shift 1 · Q54

Dual Nature of Radiation question

2022 · 27 Jun · Shift 1 · Q54

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An α\alphaα particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths (λα:λC12)({\lambda _\alpha }:{\lambda _{C12}})(λα​:λC12​) is :
  1. A
    1:31:\sqrt 31:3​
  2. B
    3:1\sqrt 3 :13​:1
  3. C
    3:13:13:1
  4. D
    2:32:\sqrt 32:3​
View written solutionFree

Correct answer: B

  1. Use de-Broglie wavelength formula

For a particle,

λ=hp\lambda = \frac{h}{p}λ=ph​

If the kinetic energy is the same, then using

K=p22mK = \frac{p^2}{2m}K=2mp2​

we get

p=2mKp = \sqrt{2mK}p=2mK​

So,

λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}λ=2mK​h​

Thus for same kinetic energy, de-Broglie wavelength is inversely proportional to the square root of mass:

λ∝1m\lambda \propto \frac{1}{\sqrt{m}}λ∝m​1​
  1. Mass of the particles
  • An α\alphaα particle has mass approximately 4u4u4u.
  • A carbon-12 atom has mass 12u12u12u.
  1. Find the ratio

Since

λ∝1m,\lambda \propto \frac{1}{\sqrt{m}},λ∝m​1​, λαλC12=mC12mα=124=3\frac{\lambda_\alpha}{\lambda_{C12}} = \sqrt{\frac{m_{C12}}{m_\alpha}} = \sqrt{\frac{12}{4}} = \sqrt{3}λC12​λα​​=mα​mC12​​​=412​​=3​

Hence,

λα:λC12=3:1\lambda_\alpha : \lambda_{C12} = \sqrt{3} : 1λα​:λC12​=3​:1
  1. Check options
  • A: 1:31: \sqrt{3}1:3​ ❌
  • B: 3:1\sqrt{3}:13​:1 ✅
  • C: 3:13:13:1 ❌
  • D: 2:32: \sqrt{3}2:3​ ❌

Therefore, the correct option is B.

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