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Dual Nature of Radiation question

2022 · 28 Jul · Shift 1 · Q62
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  5. /2022 · 28 Jul · Shift 1 · Q62

Dual Nature of Radiation question

2022 · 28 Jul · Shift 1 · Q62

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The equation λ=1.227x nm\lambda=\frac{1.227}{x} \mathrm{~nm}λ=x1.227​ nm can be used to find the de-Brogli wavelength of an electron. In this equation xxx stands for : Where m=\mathrm{m}=m= mass of electron P=\mathrm{P}=P= momentum of electron K=\mathrm{K}=K= Kinetic energy of electron V=\mathrm{V}=V= Accelerating potential in volts for electron
  1. A
    mK\sqrt{\mathrm{mK}}mK​
  2. B
    P\sqrt{\mathrm{P}}P​
  3. C
    K\sqrt{\mathrm{K}}K​
  4. D
    V\sqrt{\mathrm{V}}V​
View written solutionFree

Correct answer: D

  1. For a particle, the de-Broglie wavelength is

λ=hp\lambda=\frac{h}{p}λ=ph​

For an electron accelerated through a potential difference VVV, its kinetic energy becomes

eV=p22meV=\frac{p^2}{2m}eV=2mp2​

So,

p=2meVp=\sqrt{2meV}p=2meV​

  1. Substitute this into the de-Broglie relation:

λ=h2meV\lambda=\frac{h}{\sqrt{2meV}}λ=2meV​h​

For an electron, substituting the constants h,m,eh, m, eh,m,e gives the standard result

λ=1.227V nm\lambda = \frac{1.227}{\sqrt{V}}\ \text{nm}λ=V​1.227​ nm

  1. Comparing with the given form

λ=1.227x nm\lambda=\frac{1.227}{x}\ \text{nm}λ=x1.227​ nm

we get

x=Vx=\sqrt{V}x=V​

  1. Check options:
  • A: mK\sqrt{mK}mK​ — incorrect
  • B: P\sqrt{P}P​ — incorrect
  • C: K\sqrt{K}K​ — incorrect in this standard electron formula
  • D: V\sqrt{V}V​ — correct

Therefore, the correct answer is D.

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