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Dual Nature of Radiation question

2021 · 25 Feb · Shift 2 · Q52
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  5. /2021 · 25 Feb · Shift 2 · Q52

Dual Nature of Radiation question

2021 · 25 Feb · Shift 2 · Q52

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength λelectronλproton{{{}^\lambda electron} \over {{}^\lambda proton}}λprotonλelectron​ will be :
  1. A
    1
  2. B
    1836
  3. C
    11836{1 \over {1836}}18361​
  4. D
    918
View written solutionFree

Correct answer: B

  1. Use de Broglie wavelength formula

    The de Broglie wavelength is λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}λ=ph​=mvh​ for a particle moving with speed vvv (non-relativistic case).

  2. Given condition

    Electron and proton are moving with the same speed.

    So, λe=hmev\lambda_e = \frac{h}{m_e v}λe​=me​vh​ λp=hmpv\lambda_p = \frac{h}{m_p v}λp​=mp​vh​

  3. Take the ratio

    λeλp=h/(mev)h/(mpv)=mpme\frac{\lambda_e}{\lambda_p} = \frac{h/(m_e v)}{h/(m_p v)} = \frac{m_p}{m_e}λp​λe​​=h/(mp​v)h/(me​v)​=me​mp​​

  4. Substitute the given mass relation

    mp=1836 mem_p = 1836\,m_emp​=1836me​

    Therefore, λeλp=1836 meme=1836\frac{\lambda_e}{\lambda_p} = \frac{1836\,m_e}{m_e} = 1836λp​λe​​=me​1836me​​=1836

  5. Check options

    • A: 111 ❌
    • B: 183618361836 ✅
    • C: 11836\dfrac{1}{1836}18361​ ❌
    • D: 918918918 ❌

Therefore, the correct answer is B.

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