Question

An electron of mass m with an initial velocity \(\overrightarrow{V} = V_0 \hat{i} (V_0 \gt 0)\) enters an electric field \(\overrightarrow{E} = -E_0 \;\hat{i} (E_0=constant  \gt 0)\) at t = 0. If \(\lambda_0\) is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

• A. \(\frac{\lambda_0}{\left(1+\frac{eE_0}{mV_0}t\right)}\)
• B. \(\lambda_0t\)
• C. \(\lambda_0\left(1+\frac{eE_0}{mV_0}t\right)\)
• D. \(\lambda_0\)

Answer 1

The Correct Option is A

To determine the de-Broglie wavelength of an electron at time \( t \), we start by considering the initial conditions and the forces acting on the electron.

1. The initial velocity of the electron is \( \overrightarrow{V} = V_0 \hat{i} \), and it has a de-Broglie wavelength given by: \(\lambda_0 = \frac{h}{mv_0}\).
2. The electron enters a uniform electric field \( \overrightarrow{E} = -E_0 \hat{i} \). The force on the electron due to the electric field is: \(F = eE_0\).
3. According to Newton’s second law, the acceleration \( a \) of the electron is: \(a = \frac{F}{m} = \frac{eE_0}{m}\).
4. The velocity of the electron at time \( t \) is: \(v = v_0 + at = v_0 + \frac{eE_0}{m}t\).
5. The de-Broglie wavelength at time \( t \) can then be calculated as: \(\lambda = \frac{h}{mv}\). Substituting the expression for \( v \), we get: \(\lambda = \frac{h}{m(v_0 + \frac{eE_0}{m}t)}\).
6. Thus, the de-Broglie wavelength at time \( t \) is: \(\lambda = \frac{h}{m}\cdot\frac{1}{v_0+\frac{eE_0}{m}t} = \frac{h}{mv_0}\cdot\frac{1}{1+\frac{eE_0}{mv_0}t}\).
7. So, the de-Broglie wavelength is: \(\lambda = \frac{\lambda_0}{1+\frac{eE_0}{mV_0}t}\).

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