Question

An electron (mass m) with an initial velocity
\(\vec{v}=v_0\hat{i}(v_0>0)\)
is moving in an electric field
\(\vec{E}=E_0\hat{i}(E_0>0)\)
where E₀ is constant. If at t = 0 de Broglie wavelength is
\(λ_0=\frac{ℎ}{mv_0}\)
, then its de Broglie wavelength after time t is given by

• A. \(λ_0\)
• B. \(λ_0\left(1+\frac{eE_0t}{mv_0}\right)\)
• C. \(λ_0t\)
• D. \(\frac{λ_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)

Answer 1

The Correct Option is D

To determine the de Broglie wavelength of the electron at time \( t \), we need to consider the principles of the de Broglie wavelength and the effect of the electric field on the motion of the electron.

Initially, the de Broglie wavelength is given by: 

\(\lambda_0 = \frac{h}{mv_0}\)

where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v_0 \) is its initial velocity.

The force on the electron due to the electric field \( \vec{E} = E_0\hat{i} \) is:

\(F = eE_0\)

where \( e \) is the charge of the electron.

The resulting acceleration \( a \) is given by:

\(a = \frac{F}{m} = \frac{eE_0}{m}\)

The velocity \( v \) of the electron at time \( t \) can be obtained using the equation of motion:

\(v = v_0 + at = v_0 + \frac{eE_0t}{m}\)

The de Broglie wavelength at time \( t \) is:

\(\lambda = \frac{h}{mv}\)

Substituting the expression for \( v \) into the de Broglie wavelength formula, we have:

\(\lambda = \frac{h}{m\left(v_0 + \frac{eE_0t}{m}\right)} = \frac{h}{mv_0 + eE_0t}\)

Rearranging this in terms of \( \lambda_0 \), we get:

\(\lambda = \frac{h}{mv_0} \cdot \frac{1}{1 + \frac{eE_0t}{mv_0}} = \frac{\lambda_0}{1 + \frac{eE_0t}{mv_0}}\)

Thus, the de Broglie wavelength after time \( t \) for the electron is:

\(\frac{\lambda_0}{1 + \frac{eE_0t}{mv_0}}\)

This matches the correct answer:

\(\frac{\lambda_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)

Hence, the correct answer is the option: \(\frac{\lambda_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)

Answer 2

The correct answer is (D) : \(\frac{λ_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)

\(∴a_x=\frac{eE_0}{m}\hat{i}\)
\(∴v(t)=V_0+\frac{eE_0}{m}t\)
\(∴\frac{λ_0}{λ_2}=\frac{mv}{mV_0}\)
\(=(1+\frac{eE_0t}{mV_0})\)
\(⇒λ_2=\frac{λ_0}{\left(1+\frac{eE_0t}{mV_0}\right)}\)