Question

An electron of mass $m$ and magnitude of charge$\left|e\right|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is

• A. $-\frac{\left|e\right|Et}{h}$
• B. $-\frac{h}{\left|e\right|E\sqrt{t}}$
• C. $-\frac{h}{\left|e\right|E{t}^2}$
• D. $-\frac{2h{t}^2}{\left|e\right|E}$

Answer 1

The correct option is C

$-\frac{h}{\left|e\right|E{t}^2}$

Step 1. Given data

Mass of an electron is $m$

Charge on an electron is $\left|e\right|$

Electric field is $E$

Step 2. Finding the de-Broglie wavelength

We know that, the de-Broglie wavelength,${\lambda }_D$

${\lambda }_D=\frac{h}{p}$ [Where, $h$ is the Planck constant, $p$is momentum]

${\lambda }_D=\frac{h}{mv}$ [Where, $p=mass\left(m\right)\times velocity\left(v\right)$] [$eq\left(1\right)$]

Step 3. Finding the velocity

By using the first equation of motion,

$v=u+at$ [Where, $u$$=0$ is the initial velocity, $a$is the acceleration, $t$ is the time]

$v=at$ [$eq\left(2\right)$]

By using the formula of the force, $F$

$F=qE$ [Where, $q=\left|e\right|$is the charge, $E$ is the electric field.]

$ma=qE$ [As $F=ma$]

$a=\frac{\left|e\right|E}{m}$

Putting the value of acceleration in $eq\left(2\right)$, we get

$v=\frac{\left|e\right|Et}{m}$

Step 4. Finding the rate of change of de-Broglie wavelength

Putting the value of velocity in $eq\left(1\right)$, we get

${\lambda }_D=\frac{h}{m\left({\displaystyle \frac{\left|e\right|Et}{m}}\right)}$

${\lambda }_D=\frac{h}{\left|e\right|Et}$

Now, the rate of change of the de Broglie wavelength is given by differentiating the wavelength with respect to the time

$\frac{d}{dt}\left({\lambda }_D\right)=\frac{d}{dt}\left(\frac{h}{\left|e\right|Et}\right)$

$\frac{d}{dt}\left({\lambda }_D\right)=-\frac{h}{\left|e\right|E{t}^2}$

Therefore, the rate of change of the de-Broglie wavelength is $-\frac{h}{\left|e\right|E{t}^2}$

Hence, the correct option is $\left(C\right)$.