Question

An electron of mass m with an initial velocity of \[\vec v = {v_0}\hat i\] (${v_0} > 0$) is in an electric field of $\vec E = - {E_0}\hat i$ (${E_0} > 0$ ) . It’s De Broglie wavelength initially is ${\lambda _0}$ , so what is the De Broglie wavelength of the electron after time $t$.
A) $\dfrac{{{\lambda _0}}}{{\left( {1 + \dfrac{{e{E_0}t}}{{m{v_0}}}} \right)}}$
B) ${\lambda _0}\left( {1 + \dfrac{{e{E_0}t}}{{m{v_0}}}} \right)$
C) ${\lambda _0}$
D) ${\lambda _0}t$

Answer 1

Hint: Any charged particle that exists in an electric field is bound to experience some force. Thus the charged particle will have acceleration. That means its velocity will vary from time to time.

Complete step by step solution:
What is De Broglie wavelength?
According to De Broglie everything in nature has a wave like nature. That means everything has a wavelength. For objects which are macroscopic in nature, their wavelengths are very small and so the wave nature is ignored. But for objects which are microscopic or considerably small the wavelength is comparable to the body size of this object and so the wave nature cannot be ignored.
So, according to De Broglie the wavelength of such bodies is inversely proportional to the momentum of the body. it is mathematically written as;
$\lambda $ $\alpha $ $\dfrac{1}{{mv}}$.
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$ (Here, $h$ is the Planck's constant, $m$ is the mass and $v$ is the velocity of the body.)
So according to De Broglie the initial wavelength of the electron is given by;
${\lambda _0} = \dfrac{h}{{m{v_0}}}$ (Since, ${v_0}$ is the initial velocity) (Equation: 1)
Also, we know that when a charged particle is kept in an external electric field it experiences force.
So the electron will experience a force. This force will result in the acceleration of the electron. And, so the electron’s velocity will change.
So the electric force due to the external electric field is given by;
$F = q\left| {\vec E} \right| = \left( { - e} \right)\left( { - {E_0}} \right)$
$ \Rightarrow F = e{E_0}$ (Equation: 2)(Here, e is the charge of electron and ${E_0}$ is the magnitude of the electric field)
But, force is the product of mass and acceleration of the body;
$ \Rightarrow F = ma$ (Equation: 3)
Thus from equation 2 and equation 3 we get;
$ \Rightarrow ma = e{E_0}$
$ \Rightarrow a = \dfrac{{e{E_0}}}{m}$ (Here m is the mass of electron)(Equation: 4)
But we all know that;
$v = {v_0} + at$ (Here v is the velocity of electron after a time t)
$ \Rightarrow v = {v_0} + \dfrac{{e{E_0}}}{m}t$ (From equation: 4)(Equation: 5)
Thus by putting the values in the formula we get wavelength of electron at a time t as;
$\lambda = \dfrac{h}{{mv}}$
$ \Rightarrow \lambda = \dfrac{h}{{m\left( {{v_0} + \dfrac{{e{E_0}t}}{m}} \right)}}$
$ \Rightarrow \lambda = \dfrac{h}{{m{v_0} + e{E_0}t}}$
$ \Rightarrow \lambda = \dfrac{\dfrac{h}{mv_0}}{1+\dfrac{eE_0 t}{mv_0}}$
$ \Rightarrow \lambda = \dfrac{\lambda_0}{1+\dfrac{eE_0 t}{mv_0}}$

Thus, option A is correct.

Note: For macroscopic bodies, the size of these bodies is very huge so the wavelength is not comparable to the size of the body. Hence, their wave nature is neglected however De Broglie hypothesis is true for them also.
For microscopic particles the body size is comparable to their wavelengths and so their wave nature cannot be neglected.