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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)
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B
λ0t
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C
λ0(1+eE0mV0t)
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D
λ0
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Solution

The correct option is A $$\displaystyle\frac{\lambda_0}{\left(\displaystyle 1+\frac{eE_0}{mV_0}t\right)}$$
R.E.F image
The 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, $$ \lambda = \dfrac{h}{mV} $$
$$ = \dfrac{h}{m(V_{0}+\dfrac{eE_{0}t}{m})} $$
$$ = \dfrac{h}{mV_{0}+eE_{0}t} $$
$$ = \dfrac{1}{1/\lambda _{0}+eE_{0}t/h} $$
$$ = \dfrac{\lambda .h}{h+eE_{0}\lambda _{0}t} $$
$$ = \dfrac{\lambda _{0}}{1+\dfrac{eE_{o}\lambda _{0}t}{h}} $$
$$ = \boxed{\lambda = \dfrac{\lambda _{0}}{1+(\dfrac{eE_{0}}{mV_{0}})t}} $$ 

1123356_885644_ans_a08e842334cc4d6caed95f9e3f41389c.png

Physics

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