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Dual Nature of Radiation question

2023 · 11 Apr · Shift 2 · Q51
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  5. /2023 · 11 Apr · Shift 2 · Q51

Dual Nature of Radiation question

2023 · 11 Apr · Shift 2 · Q51

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy : (Assume mp=me×1849m_{p}=m_{e} \times 1849mp​=me​×1849 )
  1. A
    1:43
  2. B
    1:62
  3. C
    2:43
  4. D
    1:30
View written solutionFree

Correct answer: A

  1. Use de-Broglie wavelength formula

For a particle,

λ=hp\lambda = \frac{h}{p}λ=ph​

For non-relativistic motion, kinetic energy is

K=p22mK = \frac{p^2}{2m}K=2mp2​

So,

p=2mKp = \sqrt{2mK}p=2mK​

Hence,

λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}λ=2mK​h​
  1. Compare proton and electron for same kinetic energy

Since both have the same kinetic energy KKK,

λ∝1m\lambda \propto \frac{1}{\sqrt{m}}λ∝m​1​

Therefore,

λpλe=memp\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}λe​λp​​=mp​me​​​

Given,

mp=1849 mem_p = 1849\, m_emp​=1849me​

So,

λpλe=11849=143\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{1}{1849}} = \frac{1}{43}λe​λp​​=18491​​=431​

Thus,

λp:λe=1:43\lambda_p : \lambda_e = 1:43λp​:λe​=1:43
  1. Match with options

Option A: 1:431:431:43 ✅

Therefore, the correct answer is A.

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