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Dual Nature of Radiation question

2021 · 17 Mar · Shift 1 · Q61
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  5. /2021 · 17 Mar · Shift 1 · Q61

Dual Nature of Radiation question

2021 · 17 Mar · Shift 1 · Q61

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
  1. A
    1c(2mE)1/2{1 \over c}{\left( {{{2m} \over E}} \right)^{1/2}}c1​(E2m​)1/2
  2. B
    1c(E2m)1/2{1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}c1​(2mE​)1/2
  3. C
    (E2m)1/2{\left( {{E \over {2m}}} \right)^{1/2}}(2mE​)1/2
  4. D
    c(2mE)1/2c{(2mE)^{1/2}}c(2mE)1/2
View written solutionFree

Correct answer: B

  1. Wavelength of the photon

For a photon, energy is related to wavelength by E=hcλpE = \frac{hc}{\lambda_p}E=λp​hc​ So, λp=hcE\lambda_p = \frac{hc}{E}λp​=Ehc​

  1. Wavelength of the electron

For the electron, assuming non-relativistic kinetic energy, E=p22mE = \frac{p^2}{2m}E=2mp2​ Hence, p=2mEp = \sqrt{2mE}p=2mE​

Using de Broglie relation, λe=hp=h2mE\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}λe​=ph​=2mE​h​

  1. Take the ratio

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

Simplifying, λeλp=Ec2mE\frac{\lambda_e}{\lambda_p} = \frac{E}{c\sqrt{2mE}}λp​λe​​=c2mE​E​

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

  1. Match with the options

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

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