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Dual Nature of Radiation question

2020 · 9 Jan · Shift 2 · Q46
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Dual Nature of Radiation question

2020 · 9 Jan · Shift 2 · Q46

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is :
  1. A
    −h∣e∣Et{{ - h} \over {\left| e \right|Et}}∣e∣Et−h​
  2. B
    −h∣e∣Et{{ - h} \over {\left| e \right|E\sqrt t }}∣e∣Et​−h​
  3. C
    −h∣e∣Et2{{ - h} \over {\left| e \right|E{t^2}}}∣e∣Et2−h​
  4. D
    ∣e∣Eth{{\left| e \right|Et} \over h}h∣e∣Et​
View written solutionFree

Correct answer: C

  1. Force on the electron

An electron in a constant electric field EEE experiences force of magnitude F=∣e∣EF = |e|EF=∣e∣E So its acceleration is constant: a=∣e∣Ema = \frac{|e|E}{m}a=m∣e∣E​

Since the electron starts from rest, its velocity at time ttt is v=at=∣e∣Emtv = at = \frac{|e|E}{m}tv=at=m∣e∣E​t

  1. Momentum at time ttt

The momentum is p=mv=m(∣e∣Emt)=∣e∣Etp = mv = m\left(\frac{|e|E}{m}t\right) = |e|Etp=mv=m(m∣e∣E​t)=∣e∣Et

  1. de-Broglie wavelength

Using de-Broglie relation, λ=hp=h∣e∣Et\lambda = \frac{h}{p} = \frac{h}{|e|Et}λ=ph​=∣e∣Eth​

  1. Rate of change of wavelength

Differentiate with respect to time: dλdt=ddt(h∣e∣Et)\frac{d\lambda}{dt} = \frac{d}{dt}\left(\frac{h}{|e|Et}\right)dtdλ​=dtd​(∣e∣Eth​)

Since hhh, ∣e∣|e|∣e∣, and EEE are constants, dλdt=h∣e∣Eddt(t−1)\frac{d\lambda}{dt} = \frac{h}{|e|E}\frac{d}{dt}(t^{-1})dtdλ​=∣e∣Eh​dtd​(t−1) dλdt=h∣e∣E(−t−2)\frac{d\lambda}{dt} = \frac{h}{|e|E}(-t^{-2})dtdλ​=∣e∣Eh​(−t−2) dλdt=−h∣e∣Et2\frac{d\lambda}{dt} = -\frac{h}{|e|Et^2}dtdλ​=−∣e∣Et2h​

  1. Matching with options

This matches: C: −h∣e∣Et2\boxed{\text{C: } -\frac{h}{|e|Et^2}}C: −∣e∣Et2h​​

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