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Dual Nature of Radiation question

2021 · 20 Jul · Shift 2 · Q55
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Dual Nature of Radiation question

2021 · 20 Jul · Shift 2 · Q55

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron having de-Broglie wavelength λ\lambdaλ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is :
  1. A
    0
  2. B
    2m2c2λ2h2{{2{m^2}{c^2}{\lambda ^2}} \over {{h^2}}}h22m2c2λ2​
  3. C
    2mcλ2h{{2mc{\lambda ^2}} \over h}h2mcλ2​
  4. D
    hcmc{{hc} \over {mc}}mchc​
View written solutionFree

Correct answer: C

  1. Use de-Broglie relation for the incident electron

For the electron,

λ=hp\lambda = \frac{h}{p}λ=ph​

So its momentum is

p=hλp = \frac{h}{\lambda}p=λh​
  1. Find the kinetic energy of the electron

Since the options are based on the non-relativistic result, use

K=p22mK = \frac{p^2}{2m}K=2mp2​

Substituting p=hλp = \dfrac{h}{\lambda}p=λh​,

K=12m(hλ)2=h22mλ2K = \frac{1}{2m}\left(\frac{h}{\lambda}\right)^2 = \frac{h^2}{2m\lambda^2}K=2m1​(λh​)2=2mλ2h2​
  1. Condition for cut-off wavelength in X-ray tube

The minimum (cut-off) wavelength occurs when the entire kinetic energy of the electron is converted into a single photon:

K=hν=hcλmin⁡K = h\nu = \frac{hc}{\lambda_{\min}}K=hν=λmin​hc​

Therefore,

λmin⁡=hcK\lambda_{\min} = \frac{hc}{K}λmin​=Khc​

Substitute K=h22mλ2K = \dfrac{h^2}{2m\lambda^2}K=2mλ2h2​:

λmin⁡=hch2/(2mλ2)=hc⋅2mλ2h2=2mcλ2h\lambda_{\min} = \frac{hc}{h^2/(2m\lambda^2)} = hc \cdot \frac{2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}λmin​=h2/(2mλ2)hc​=hc⋅h22mλ2​=h2mcλ2​
  1. Match with the options
λmin⁡=2mcλ2h\lambda_{\min} = \frac{2mc\lambda^2}{h}λmin​=h2mcλ2​

This matches Option C.

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So, they agree.

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