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An electron of mass m with an initial velocity $\vec{V} = V_0\hat{i}\ (V_0 > 0)$ enters an electric field $\vec{E} = -E_0\hat{i}$ ($E_0$ = constant > 0) at t = 0. If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is
Explanation
The electron speeds up, so $\lambda = h/mV$ decreases as $\lambda_0/(1 + eE_0t/mV_0)$.
Detailed Solution
Initial de-Broglie wavelength $\lambda_0 = \frac{h}{mV_0}$ ...(i)
Acceleration of electron $a = \frac{eE_0}{m}$ (along +x, since the field is along –x)
Velocity after time t: $V = V_0 + \frac{eE_0}{m}t$
So, $\lambda = \frac{h}{mV} = \frac{h}{m\left(V_0 + \frac{eE_0}{m}t\right)} = \frac{h}{mV_0\left[1 + \frac{eE_0}{mV_0}t\right]}$ ...(ii)
Divide (ii) by (i): $\lambda = \frac{\lambda_0}{\left[1 + \frac{eE_0}{mV_0}t\right]}$
Acceleration of electron $a = \frac{eE_0}{m}$ (along +x, since the field is along –x)
Velocity after time t: $V = V_0 + \frac{eE_0}{m}t$
So, $\lambda = \frac{h}{mV} = \frac{h}{m\left(V_0 + \frac{eE_0}{m}t\right)} = \frac{h}{mV_0\left[1 + \frac{eE_0}{mV_0}t\right]}$ ...(ii)
Divide (ii) by (i): $\lambda = \frac{\lambda_0}{\left[1 + \frac{eE_0}{mV_0}t\right]}$