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

Answer: (B)

$\frac{\lambda_{0}}{\left(1+ \frac{eE_{0}}{mV_{0}} t\right)}$

Answer 2

Initial de-Broglie wavelength
$\lambda_0 = \frac{h}{mV_0}$ ....(i)

Acceleration of electron
$a = \frac{e E_0}{m}$
Velocity after time ? $t$ ?
$V = \left(V_{0} + \frac{eE_{0}}{m} t\right)$
So, $\lambda = \frac{h}{mV} = \frac{h}{m\left(V_{0} + \frac{eE_{0}}{m}t\right)}$
$= \frac{h}{mV_{0} \left[1+ \frac{eE_{0}}{mV_{0}}t\right]} = \frac{\lambda_{0}}{\left[1+ \frac{eE_{0}}{mV_{0}}t\right]}$ ....(ii)
Divide (ii) by (i),
$\lambda = \frac{\lambda_{0}}{\left[1+ \frac{eE_{0}}{mV_{0}}t \right]}$