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Dual Nature of Radiation question

2020 · 8 Jan · Shift 2 · Q41
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  5. /2020 · 8 Jan · Shift 2 · Q41

Dual Nature of Radiation question

2020 · 8 Jan · Shift 2 · Q41

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron (mass m) with initial velocity v→=v0i^+v0j^\overrightarrow v = {v_0}\widehat i + {v_0}\widehat jv=v0​i+v0​j​ is in an electric field E→=−E0k^\overrightarrow E = - {E_0}\widehat kE=−E0​k. If λ0\lambda _0λ0​ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :
  1. A
    λ01+e2E2t2m2v02{{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}1+m2v02​e2E2t2​​λ0​​
  2. B
    λ021+e2E2t2m2v02{{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}1+m2v02​e2E2t2​​λ0​2​​
  3. C
    λ01+e2E2t22m2v02{{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}1+2m2v02​e2E2t2​​λ0​​
  4. D
    λ02+e2E2t2m2v02{{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}2+m2v02​e2E2t2​​λ0​​
View written solutionFree

Correct answer: C

  1. Initial velocity and momentum

The electron has initial velocity

v⃗(0)=v0i^+v0j^\vec v(0)=v_0\hat i+v_0\hat jv(0)=v0​i^+v0​j^​

So its initial speed is

vi=v02+v02=2 v0v_i=\sqrt{v_0^2+v_0^2}=\sqrt{2}\,v_0vi​=v02​+v02​​=2​v0​

Hence the initial de-Broglie wavelength is

λ0=hmvi=hm2v0\lambda_0=\frac{h}{m v_i}=\frac{h}{m\sqrt{2}v_0}λ0​=mvi​h​=m2​v0​h​
  1. Acceleration due to electric field

Given electric field:

E⃗=−E0k^\vec E=-E_0\hat kE=−E0​k^

Force on electron (−e)(-e)(−e) is

F⃗=qE⃗=(−e)(−E0k^)=eE0k^\vec F = q\vec E = (-e)(-E_0\hat k)=eE_0\hat kF=qE=(−e)(−E0​k^)=eE0​k^

Therefore acceleration is

a⃗=F⃗m=eE0mk^\vec a=\frac{\vec F}{m}=\frac{eE_0}{m}\hat ka=mF​=meE0​​k^

So only the kkk-component of velocity changes with time.

  1. Velocity at time ttt

Since there is no acceleration in xxx and yyy directions,

vx=v0,vy=v0v_x=v_0,\qquad v_y=v_0vx​=v0​,vy​=v0​

And along zzz,

vz=eE0mtv_z=\frac{eE_0}{m}tvz​=meE0​​t

Thus,

v⃗(t)=v0i^+v0j^+eE0tmk^\vec v(t)=v_0\hat i+v_0\hat j+\frac{eE_0 t}{m}\hat kv(t)=v0​i^+v0​j^​+meE0​t​k^
  1. Speed at time ttt

The magnitude of velocity is

v(t)=v02+v02+(eE0tm)2v(t)=\sqrt{v_0^2+v_0^2+\left(\frac{eE_0 t}{m}\right)^2}v(t)=v02​+v02​+(meE0​t​)2​ =2v02+e2E02t2m2=\sqrt{2v_0^2+\frac{e^2E_0^2 t^2}{m^2}}=2v02​+m2e2E02​t2​​
  1. De-Broglie wavelength at time ttt

Using

λ=hmv(t)\lambda=\frac{h}{mv(t)}λ=mv(t)h​

we get

λ(t)=hm2v02+e2E02t2m2\lambda(t)=\frac{h}{m\sqrt{2v_0^2+\frac{e^2E_0^2 t^2}{m^2}}}λ(t)=m2v02​+m2e2E02​t2​​h​

Factor out 2v022v_0^22v02​ from the square root:

λ(t)=hm2v01+e2E02t22m2v02\lambda(t)=\frac{h}{m\sqrt{2}v_0\sqrt{1+\frac{e^2E_0^2 t^2}{2m^2v_0^2}}}λ(t)=m2​v0​1+2m2v02​e2E02​t2​​h​

But

λ0=hm2v0\lambda_0=\frac{h}{m\sqrt{2}v_0}λ0​=m2​v0​h​

Therefore,

λ(t)=λ01+e2E02t22m2v02\boxed{\lambda(t)=\frac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2 t^2}{2m^2v_0^2}}}}λ(t)=1+2m2v02​e2E02​t2​​λ0​​​
  1. Match with options

This matches Option C:

λ01+e2E2t22m2v02\boxed{\frac{\lambda_0}{\sqrt{1+\frac{e^2E^2 t^2}{2m^2v_0^2}}}}1+2m2v02​e2E2t2​​λ0​​​
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