Single Choice

An electron (mass $$m$$) with initial velocity $$\vec { v } ={ v }_{ 0 }\hat { i } +{ v }_{ 0 }\hat { j } $$ is an electric field $$\vec { E } =-{ E }_{ 0 }\hat { k } $$. If $${ \lambda }_{ 0 }$$ is initial de-Broglie wave length at time $$t$$ is given by

A$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ 2{ m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$
Correct Answer
B$$\cfrac { { \lambda }_{ 0 }\sqrt { 2 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$
C$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 2+\cfrac { { e }^{ 2 }{ E }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$
D$$\cfrac { { \lambda }_{ 0 } }{ \sqrt { 1+\cfrac { { e }^{ 2 }{ E }_{ 0 }^{ 2 }{ t }^{ 2 } }{ { m }^{ 2 }{ v }_{ 0 }^{ 2 } } } } $$

Solution

Initially momentum, $$P=\dfrac{h}{\lambda_0}$$ $$\because V_i = \sqrt{V_0^2 + V_0^2} = \sqrt{2}V_0$$ $$m(\sqrt{2}V_0) = \dfrac{h}{\lambda_0}$$ Now, velocity as a function of time $$= V_0\hat{i} + V_0\hat{j} + \dfrac{eE_0}{m} + \hat{k}$$ because only force is acting in $$Z-dir^n$$. So velocity will be change in $$Z.dir^n$$ and keeping constant in $$x-y \ dir^n$$. $$\therefore V_{net} = \sqrt{V_0^2+V_0^2 + \left(\left( \dfrac{eE_0}{m}\right)\right)^2}$$ $$\Rightarrow P = \dfrac{h}{\lambda}$$ $$\Rightarrow mV_{net} = \dfrac{h}{\lambda}$$ $$\Rightarrow \lambda = \dfrac{h}{mV_{net}} = \dfrac{h}{m\sqrt{2V_0^2 + \dfrac{e^2E_0^2}{m^2}t^2}}$$ $$\left( \because \lambda_0 = \dfrac{h}{m\sqrt{2}V_0}\right)$$ $$\Rightarrow \lambda = \dfrac{h}{\sqrt{\left(1 + \dfrac{e^2E_0^2}{2m^2V_0^2}t^2\right)}}$$

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