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Dual Nature of Radiation question

2022 · 27 Jul · Shift 1 · Q61
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  5. /2022 · 27 Jul · Shift 1 · Q61

Dual Nature of Radiation question

2022 · 27 Jul · Shift 1 · Q61

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron (mass m\mathrm{m}m) with an initial velocity v⃗=v0i^(v0>0)\vec{v}=v_{0} \hat{i}\left(v_{0}\gt 0\right)v=v0​i^(v0​>0) is moving in an electric field E⃗=−E0i^(E0>0)\vec{E}=-E_{0} \hat{i}\left(E_{0}\gt 0\right)E=−E0​i^(E0​>0) where E0E_{0}E0​ is constant. If at t=0\mathrm{t}=0t=0 de Broglie wavelength is λ0=hmv0\lambda_{0}=\frac{h}{m v_{0}}λ0​=mv0​h​, then its de Broglie wavelength after time t is given by
  1. A
    λ0\lambda_{0}λ0​
  2. B
    λ0(1+eE0tmv0)\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)λ0​(1+mv0​eE0​t​)
  3. C
    λ0t\lambda_{0} tλ0​t
  4. D
    λ0(1+eE0tmv0)\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}(1+mv0​eE0​t​)λ0​​
View written solutionFree

Correct answer: D

  1. Force on the electron

The electric field is

E⃗=−E0i^\vec E = -E_0\hat iE=−E0​i^

with E0>0E_0>0E0​>0.

For an electron, charge is

q=−eq=-eq=−e

So the force is

F⃗=qE⃗=(−e)(−E0i^)=eE0i^\vec F = q\vec E = (-e)(-E_0\hat i)= eE_0\hat iF=qE=(−e)(−E0​i^)=eE0​i^

Thus the electron experiences a force in the +x+x+x direction.

  1. Acceleration of the electron

Using F⃗=ma⃗\vec F = m\vec aF=ma,

a⃗=eE0mi^\vec a = \frac{eE_0}{m}\hat ia=meE0​​i^

So the acceleration is constant and along the direction of initial velocity.

  1. Velocity after time ttt

Initial velocity is

v⃗(0)=v0i^\vec v(0)=v_0\hat iv(0)=v0​i^

Hence after time ttt,

v⃗(t)=(v0+eE0mt)i^\vec v(t)=\left(v_0+\frac{eE_0}{m}t\right)\hat iv(t)=(v0​+meE0​​t)i^

So the speed is

v(t)=v0+eE0mtv(t)=v_0+\frac{eE_0}{m}tv(t)=v0​+meE0​​t
  1. de Broglie wavelength

The de Broglie wavelength is

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}λ=ph​=mvh​

Therefore at time ttt,

λ(t)=hm(v0+eE0mt)\lambda(t)=\frac{h}{m\left(v_0+\frac{eE_0}{m}t\right)}λ(t)=m(v0​+meE0​​t)h​

Simplify:

λ(t)=hmv0+eE0t\lambda(t)=\frac{h}{mv_0+eE_0 t}λ(t)=mv0​+eE0​th​

Now given

λ0=hmv0\lambda_0=\frac{h}{mv_0}λ0​=mv0​h​

so

h=λ0mv0h = \lambda_0 m v_0h=λ0​mv0​

Substitute into the expression:

λ(t)=λ0mv0mv0+eE0t\lambda(t)=\frac{\lambda_0 m v_0}{m v_0+eE_0 t}λ(t)=mv0​+eE0​tλ0​mv0​​

Factor mv0mv_0mv0​ from the denominator:

λ(t)=λ01+eE0tmv0\lambda(t)=\frac{\lambda_0}{1+\frac{eE_0 t}{m v_0}}λ(t)=1+mv0​eE0​t​λ0​​
  1. Match with the options

This matches:

λ0(1+eE0tmv0)\boxed{\frac{\lambda_0}{\left(1+\frac{eE_0 t}{m v_0}\right)}}(1+mv0​eE0​t​)λ0​​​

So the correct option is D.

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