An electron (mass m ) with an initial velocity → v = v _ 0 ^ i ( v _ 0 0) is moving in an electric field → E = - E _ 0 ^ i ( E _ 0 0) where E _ 0 is constant. If at t = 0 , de-Broglie wavelength is λ _ 0 = h m v _ 0

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An electron (mass m ) with an initial velocity → v = v _ 0 ^ i ( v _ 0 > 0) is moving in an electric field → E = - E _ 0 ^ i ( E _ 0 > 0) where E _ 0 is constant. If at t = 0 , de-Broglie wavelength is λ _ 0 = h m v _ 0

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Correct answer: D. λ _ 0 (1 + e E _ 0 t m v _ 0 )

Initial de-Broglie wavelength of electron,
${λ}_{0} = \dfrac{h}{m{v}_{0}}$ ....(i)
Force on electron in electric field,
$F = - eE = - e\left[- {E}_{0}\overset{^}{i}\right] = e{E}_{0}\overset{^}{i}$
Acceleration of electron, $a = \dfrac{F}{m} = \dfrac{e{E}_{0}\overset{^}{i}}{m}$
Velocity of electron after time $t$ ,
$v = {v}_{0}\overset{^}{i} + \left(\dfrac{e{E}_{0}\overset{^}{i}}{m}\right)t = \left({v}_{0} + \dfrac{e{E}_{0}}{m}t\right)\overset{^}{i} = {v}_{0}\left(1 + \dfrac{e{E}_{0}}{m{v}_{0}}t\right)\overset{^}{i}$
de-Broglie wavelength associated with electron at time $t$ is
$λ = \dfrac{h}{mv}$
$\to λ = \dfrac{h}{m\left[{v}_{0}\left(1 + \dfrac{e{E}_{0}}{m{v}_{0}}t\right)\right]} = \dfrac{{λ}_{0}}{\left[1 + \dfrac{e{E}_{0}}{m{v}_{0}}t\right]}$

An electron (mass m ) with an initial velocity → v = v _ 0 ^ i ( v...

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