An electron (mass m) with an initial velocity | vedclass

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(D) The force on the electron in the electric field is $\overrightarrow{F} = q\overrightarrow{E} = (-e)(-E_{0} \hat{i}) = eE_{0} \hat{i}$.Since the fo

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$\frac{\lambda_{0}}{\left(1 + \frac{e E_{0} t}{mv_{0}}\right)}$

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(D) The force on the electron in the electric field is $\overrightarrow{F} = q\overrightarrow{E} = (-e)(-E_{0} \hat{i}) = eE_{0} \hat{i}$.Since the force is in the direction of the initial velocity,the electron undergoes uniform acceleration $a = \frac{F}{m} = \frac{eE_{0}}{m}$.The velocity of the electron at time $t$ is $v(t) = v_{0} + at = v_{0} + \frac{eE_{0}t}{m}$.The de Broglie wavelength at time $t$ is given by $\lambda = \frac{h}{mv(t)}$.Substituting the expression for $v(t)$,we get $\lambda = \frac{h}{m(v_{0} + \frac{eE_{0}t}{m})} = \frac{h}{mv_{0} + eE_{0}t}$.Factoring out $mv_{0}$ in the denominator,we get $\lambda = \frac{h}{mv_{0}(1 + \frac{eE_{0}t}{mv_{0}})}$.Since $\lambda_{0} = \frac{h}{mv_{0}}$,we can write $\lambda = \frac{\lambda_{0}}{1 + \frac{eE_{0}t}{mv_{0}}}$.

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