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Dual Nature of Radiation question

2021 · 24 Feb · Shift 2 · Q59
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  5. /2021 · 24 Feb · Shift 2 · Q59

Dual Nature of Radiation question

2021 · 24 Feb · Shift 2 · Q59

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The de-Broglie wavelength of a proton and α\alphaα-particle are equal. The ratio of their velocities is :
  1. A
    4 : 2
  2. B
    4 : 3
  3. C
    4 : 1
  4. D
    1 : 4
View written solutionFree

Correct answer: C

  1. Use de-Broglie relation

For any particle,

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}λ=ph​=mvh​

for non-relativistic motion.

Since the de-Broglie wavelengths of the proton and the α\alphaα-particle are equal,

λp=λα\lambda_p = \lambda_\alphaλp​=λα​

So,

hmpvp=hmαvα\frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha}mp​vp​h​=mα​vα​h​

Cancelling hhh,

mpvp=mαvαm_p v_p = m_\alpha v_\alphamp​vp​=mα​vα​
  1. Use mass of α\alphaα-particle

An α\alphaα-particle has mass approximately equal to 444 times the mass of a proton:

mα≈4mpm_\alpha \approx 4m_pmα​≈4mp​

Substitute into the momentum equality:

mpvp=4mpvαm_p v_p = 4m_p v_\alphamp​vp​=4mp​vα​

Cancelling mpm_pmp​,

vp=4vαv_p = 4v_\alphavp​=4vα​

Therefore,

vp:vα=4:1v_p : v_\alpha = 4:1vp​:vα​=4:1
  1. Check options
  • A: 4:2=2:14:2 = 2:14:2=2:1 ❌
  • B: 4:34:34:3 ❌
  • C: 4:14:14:1 ✅
  • D: 1:41:41:4 ❌

Hence, the correct option is C.

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