Question

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} (V_0 > 0)$ enters an electric field $\vec{E} = - E_0 \hat{i} (E_0 = \text{constant } > 0)$ at $t = 0$. If $\lambda_o$ is its de-Broglie wavelength initially,. then its de-Broglie wavelength at time $t$ is

• A. $\lambda_0$
• B. $\frac{\lambda_{0}}{\left(1+ \frac{eE_{0}}{mV_{0}} t\right)} $
• C. $\lambda_0 \, t $
• D. $\lambda_0 \left(1+ \frac{eE_{0}}{mV_{0}} t\right)$

Answer 1

The Correct Option is B

To solve this problem, we need to find the change in the de-Broglie wavelength of an electron as it moves through an electric field.

Step 1: Initial Understanding

An electron with an initial velocity V_0 enters a uniform electric field \vec{E} = -E_0 \hat{i}. The force acting on the electron due to the electric field is given by:

F = -eE_0

Since force is the rate of change of momentum, we have:

\frac{dp}{dt} = -eE_0

Step 2: Calculate the momentum at time t

The initial momentum p_0 of the electron is given by:

p_0 = mV_0

The change in momentum after time t is:

p = p_0 - eE_0 t

Substitute initial momentum into the above equation:

p = mV_0 - eE_0 t

Step 3: Find the new de-Broglie wavelength

The de-Broglie wavelength \lambda of a particle is given by:

\lambda = \frac{h}{p}

Initial de-Broglie wavelength \lambda_0 is:

\lambda_0 = \frac{h}{mV_0}

The new de-Broglie wavelength at time t is:

\lambda = \frac{h}{mV_0 - eE_0 t}

Using the expression for \lambda_0, we can express the new wavelength \lambda as:

\lambda = \frac{\lambda_0}{1 + \frac{eE_0}{mV_0} t}

Thus, the de-Broglie wavelength of the electron at time t is \frac{\lambda_0}{\left(1+ \frac{eE_0}{mV_0} t\right)}.

Conclusion: The correct answer is \frac{\lambda_0}{\left(1+ \frac{eE_0}{mV_0} t\right)}.