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Dual Nature of Radiation question

2020 · 7 Jan · Shift 2 · Q58
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  5. /2020 · 7 Jan · Shift 2 · Q58

Dual Nature of Radiation question

2020 · 7 Jan · Shift 2 · Q58

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)
  1. A
    1c(2Em)12{1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}}c1​(m2E​)21​
  2. B
    1c(E2m)12{1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}c1​(2mE​)21​
  3. C
    (E2m)12{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}(2mE​)21​
  4. D
    c(2mE)12c{\left( {2mE} \right)^{{1 \over 2}}}c(2mE)21​
View written solutionFree

Correct answer: B

  1. Given:

    • Electron energy =E= E=E
    • Photon energy =E= E=E
    • We need λeλp\frac{\lambda_e}{\lambda_p}λp​λe​​ where λe\lambda_eλe​ is the de-Broglie wavelength of the electron and λp\lambda_pλp​ is the wavelength of the photon.

  2. Photon wavelength

    For a photon, E=hcλpE = \frac{hc}{\lambda_p}E=λp​hc​ so λp=hcE\lambda_p = \frac{hc}{E}λp​=Ehc​

  3. Electron de-Broglie wavelength

    Since energy is in the range of a few eV, the electron is non-relativistic.

    Its kinetic energy is E=p22mE = \frac{p^2}{2m}E=2mp2​ Therefore, p=2mEp = \sqrt{2mE}p=2mE​

    De-Broglie wavelength of electron: λe=hp=h2mE\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}λe​=ph​=2mE​h​

  4. Take the ratio

    λeλp=h2mEhcE\frac{\lambda_e}{\lambda_p} = \frac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}λp​λe​​=Ehc​2mE​h​​

    Cancel hhh: λeλp=12mE⋅Ec\frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{2mE}}\cdot \frac{E}{c}λp​λe​​=2mE​1​⋅cE​

    λeλp=1c⋅E2mE\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\cdot \frac{E}{\sqrt{2mE}}λp​λe​​=c1​⋅2mE​E​

    Simplify: E2mE=E2m\frac{E}{\sqrt{2mE}} = \sqrt{\frac{E}{2m}}2mE​E​=2mE​​

    Hence, λeλp=1cE2m\frac{\lambda_e}{\lambda_p} = \frac{1}{c}\sqrt{\frac{E}{2m}}λp​λe​​=c1​2mE​​

  5. Match with options

    This corresponds to Option B: 1c(E2m)1/2\boxed{\frac{1}{c}\left(\frac{E}{2m}\right)^{1/2}}c1​(2mE​)1/2​

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