Dual Nature of Matter

JEE Main level, single correct, medium

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Question 7 of 200

An electron of mass mm is moving in an electric field E⃗=−2E0i^\vec{E} = -2E_0\hat{i} (E0=E_0 = constant >0\gt 0), with an initial velocity V⃗=v0i^\vec{V} = v_0\hat{i} (v0=v_0 = constant >0\gt 0). If λ0=h4mv0\lambda_0 = \dfrac{h}{4mv_0}, its de Broglie wavelength at time tt is __________.
(e=e = charge of electron)

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Answer: (C) 4λ0[1+2E0emtv0]\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}

The force experienced by the electron in the electric field is given by F⃗=qE⃗\vec{F} = q\vec{E}.

Since the charge of an electron is −e-e and the electric field is E⃗=−2E0i^\vec{E} = -2E_0\hat{i}, the force is:

F⃗=(−e)(−2E0i^)=2eE0i^\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}

The acceleration of the electron is:

a⃗=F⃗m=2eE0mi^\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}

Using the first equation of motion, the velocity of the electron at time tt is:

v⃗(t)=V⃗+a⃗t=(v0+2eE0mt)i^\vec{v}(t) = \vec{V} + \vec{a}t = \left(v_0 + \dfrac{2eE_0}{m}t\right)\hat{i}

The de Broglie wavelength of the electron at time tt is:

λ=hm∣v⃗(t)∣=hm(v0+2eE0mt)\lambda = \dfrac{h}{m|\vec{v}(t)|} = \dfrac{h}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}

We are given that λ0=h4mv0\lambda_0 = \dfrac{h}{4mv_0}, which implies h=4mv0λ0h = 4mv_0\lambda_0. Substituting this into the expression for λ\lambda:

λ=4mv0λ0m(v0+2eE0mt)\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}

Dividing the numerator and the denominator by mv0mv_0, we get:

λ=4λ01+2E0emtv0\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}

Answer: 4λ0[1+2E0emtv0]\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}

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