Question

An electron of mass \( m \) with initial velocity \( \vec{v} = v_0 \hat{i} \) (\( v_0>0 \)) enters in an electric field \( \vec{E} = -E_0 \hat{i} \) (\( E_0 \) is constant \(>0 \)) at \( t = 0 \). If \( \lambda \) is its de-Broglie wavelength initially, then the de-Broglie wavelength after time \( t \) is:

• A. \( \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \)
• B. \( \frac{\lambda}{\left(1 - \frac{eE_0 t}{m v_0}\right)^2} \)
• C. \( \left(1 + \frac{eE_0 t}{m v_0}\right) \lambda \)
• D. \( \left(1 + \frac{eE_0 t}{m v_0}\right)^2 \lambda \)

Hint:

For charged particles moving in an electric field, use Newton’s second law to determine velocity change over time. The de-Broglie wavelength is inversely proportional to momentum, allowing derivation of the new wavelength using \( \lambda = \frac{h}{mv} \).

Answer 1

The Correct Option is A

To solve the problem of determining the change in the de Broglie wavelength of an electron in an electric field, we start by analyzing its motion under the field's influence.

Initial Conditions:

Initial velocity of the electron: \( \vec{v} = v_0 \hat{i} \)

Initial de Broglie wavelength: \[ \lambda = \frac{h}{mv_0} \] where \( h \) is Planck’s constant, \( m \) is the mass of the electron, and \( v_0 \) is the initial velocity.

Effect of Electric Field:

The electron enters a uniform electric field: \( \vec{E} = -E_0 \hat{i} \)

Force on the electron: \[ \vec{F} = e\vec{E} = -eE_0 \hat{i} \] Acceleration: \[ a = \frac{F}{m} = -\frac{eE_0}{m} \]

Velocity After Time \( t \):

\[ v(t) = v_0 + at = v_0 - \frac{eE_0 t}{m} \]

New de Broglie Wavelength:

\[ \lambda'(t) = \frac{h}{m v(t)} = \frac{h}{m \left( v_0 - \frac{eE_0 t}{m} \right)} \]

Simplified Expression:

Factor and simplify: \[ \lambda'(t) = \frac{h}{mv_0 \left(1 - \frac{eE_0 t}{mv_0} \right)} = \frac{\lambda}{1 + \frac{eE_0 t}{mv_0}} \]

Conclusion:

The de Broglie wavelength of the electron at time \( t \) in the electric field is: \[ \boxed{ \lambda'(t) = \frac{\lambda}{1 + \frac{eE_0 t}{mv_0}} } \]

Answer 2

Step 1: Equation of Motion for the Electron The force on the electron due to the electric field is given by: \[ F = eE_0 \] Using Newton’s second law: \[ m \frac{dv}{dt} = -eE_0 \] Integrating both sides from \( v_0 \) to \( v \) over time \( 0 \) to \( t \): \[ v = v_0 - \frac{eE_0 t}{m} \] Step 2: De-Broglie Wavelength Relation The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{mv} \] After time \( t \), the new wavelength \( \lambda' \) is: \[ \lambda' = \frac{h}{m v} \] Substituting \( v = v_0 - \frac{eE_0 t}{m} \): \[ \lambda' = \frac{h}{m \left( v_0 - \frac{eE_0 t}{m} \right)} \] Dividing the numerator and denominator by \( h / m v_0 \), we get: \[ \lambda' = \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \] Step 3: Conclusion Thus, the de-Broglie wavelength after time \( t \) is \( \frac{\lambda}{1 + \frac{eE_0 t}{m v_0}} \), which matches option (1).