Question

An electron of mass 'm' and charge 'e' initially at rest gets accelerated by a constant electric field 'E'. The rate of change of de-Broglie wavelength of the electron at time 't' is (Ignore relativistic effect) ($h$ = Planck's constant) ______.

• A. $-\frac{h}{e E t^2}$
• B. $-\frac{e E t}{h}$
• C. $-\frac{m h}{e E t^2}$
• D. $-\frac{h}{e E}$

Hint:

It's fascinating to note that the rate of change of the de-Broglie wavelength in a uniform electric field is completely independent of the mass of the particle! Only its charge matters.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An electron is undergoing constant acceleration due to a uniform electric field. As it speeds up, its de-Broglie wavelength shrinks. We need to find the instantaneous rate of change of this wavelength with respect to time ($\frac{d\lambda}{dt}$).

Step 2: Key Formula or Approach:
1. Force on an electron: $F = eE$.
2. Acceleration (Newton's 2nd Law): $a = \frac{F}{m} = \frac{eE}{m}$.
3. Velocity at time $t$ (Kinematics, starting from rest): $v = at = \frac{eE}{m}t$.
4. de-Broglie wavelength: $\lambda = \frac{h}{mv}$.
5. Differentiate $\lambda$ with respect to $t$.

Step 3: Detailed Explanation:
Substitute the velocity expression into the de-Broglie wavelength formula:
$$\lambda = \frac{h}{m \left( \frac{eE}{m}t \right)}$$
The mass '$m$' perfectly cancels out!
$$\lambda = \frac{h}{eEt}$$
Now, take the derivative of $\lambda$ with respect to time $t$:
$$\frac{d\lambda}{dt} = \frac{d}{dt} \left( \frac{h}{eE} \cdot t^{-1} \right)$$
Since $\frac{h}{eE}$ is entirely constant, we just differentiate $t^{-1}$ using the power rule:
$$\frac{d\lambda}{dt} = \frac{h}{eE} \left( -1 \cdot t^{-2} \right)$$
$$\frac{d\lambda}{dt} = -\frac{h}{e E t^2}$$
The negative sign correctly indicates that as time goes on and the electron accelerates, its wavelength is decreasing.

Step 4: Final Answer:
The rate of change is $-\frac{h}{e E t^2}$, matching option (a).