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Question: An electron (mass m) with an initial velocity \[\bar v = {v_0}\hat i\] is an electric field \[\bar E...

An electron (mass m) with an initial velocity vˉ=v0i^\bar v = {v_0}\hat i is an electric field Eˉ=E0j^\bar E = {E_0}\hat j. If λ0=hmv0{\lambda _0} = \dfrac{h}{{m{v_0}}}, its de Broglie wavelength at time t is given by
A. λ0{\lambda _0}
B. λ01+e2E02t2m2v02{\lambda _0}\sqrt {1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}}
C. λ01+e2E02t2m2v02\dfrac{{{\lambda _0}}}{{\sqrt {1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}} }}
D. λ01+e2E02t2m2v02\dfrac{{{\lambda _0}}}{{1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}}}

Explanation

Solution

To solve this question we have to know about de Broglie’s equation properly. This equation describes the wave nature of a particle. The wavelength of any moving object or particle is proportional to the product of mass of the particle and the velocity of the particle. Here the constant is h. here we know that h is equal to the Planck’s constant.

Complete step by step solution:
We know that, initial de Broglie wavelength of electron λ0=hmv0{\lambda _0} = \dfrac{h}{{m{v_0}}} is given in the question.Now, we also know that the force on the electron in electric field is, Fˉ=−eEˉ=−eE0j^\bar F = - e\bar E = - e{E_0}\hat j
Acceleration of electrons is, aˉ=Fm=eE0mj^\bar a = \dfrac{F}{m} = \dfrac{{e{E_0}}}{m}\hat j
This is acting along the negative y axis. We know, Initial velocity of electron along x axis is vˉx0=v0i^{\bar v_{x0}} = {v_0}\hat i
Now, according to the question, we can consider, the initial velocity of the electron along y axis is zero.Now, velocity of electron after time t along x axis is equal to v0i^{v_0}\hat i
Velocity of electron along y axis after t time is 0+(−eE0mj^)t=−eE0mtj^0 + ( - \dfrac{{e{E_0}}}{m}\hat j)t = - \dfrac{{e{E_0}}}{m}t\hat j
We know that, magnitude of velocity of electron after time t is,
\left| {\bar v} \right| = \sqrt {{v_x}^2 + {v_y}^2} = \sqrt {{v_0}^2 + {{( - \dfrac{{e{E_0}}}{m}t)}^2}} $$$$ = {v_0}\sqrt {1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}}
Now, de Broglie wavelength at time t is equal to,

\Rightarrow\lambda = \dfrac{h}{{m{v_0}\sqrt {1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}} }} \\\ \therefore\lambda = \dfrac{{{\lambda _0}}}{{\sqrt {1 + \dfrac{{{e^2}{E_0}^2{t^2}}}{{{m^2}{v_0}^2}}} }}$$ This is the correct final answer. **So, option C is correct.** **Note:** We can get confused between the velocity along x axis and the velocity along y axis. Here according to this question, the initial velocity of the electron along y axis is zero. But, there is a value of the velocity along y axis after t time, and the value of velocity along x axis after t time is $${v_0}$$.