Question

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

A
λ0(1+eE0mV0t)
B
λ0t
C
λ0(1+eE0mV0t)
D
λ0

Solution

## The correct option is A $$\displaystyle\frac{\lambda_0}{\left(\displaystyle 1+\frac{eE_0}{mV_0}t\right)}$$R.E.F imageThe diagram will be :- Initial velocity $$= V_{0}$$$$\because$$ Acceleration is constant :-Final velocity $$= v = v_{0}+\dfrac{eE_{0}}{m}t$$(at time t)Now, at $$t = 0 \Rightarrow$$ de-Broglie wavelength $$= \lambda _{0}$$$$\Rightarrow \lambda _{0} = \dfrac{h}{mv_{0}}$$Let at time t, de-Broglie wavelength $$= \lambda$$Then, $$\lambda = \dfrac{h}{mV}$$$$= \dfrac{h}{m(V_{0}+\dfrac{eE_{0}t}{m})}$$$$= \dfrac{h}{mV_{0}+eE_{0}t}$$$$= \dfrac{1}{1/\lambda _{0}+eE_{0}t/h}$$$$= \dfrac{\lambda .h}{h+eE_{0}\lambda _{0}t}$$$$= \dfrac{\lambda _{0}}{1+\dfrac{eE_{o}\lambda _{0}t}{h}}$$$$= \boxed{\lambda = \dfrac{\lambda _{0}}{1+(\dfrac{eE_{0}}{mV_{0}})t}}$$ Physics

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