A) \[\frac{{{\lambda }_{0}}}{\left( 1+\frac{e{{E}_{0}}}{m{{v}_{0}}}t \right)}\]
B) \[{{\lambda }_{0}}\left( 1+\frac{e{{E}_{0}}}{m{{v}_{0}}}t \right)\]
C) \[{{\lambda }_{0}}t\]
D) \[{{\lambda }_{0}}\]
Correct Answer: A
Solution :
[a] : Here, \[\vec{E}=-{{E}_{0}}\hat{i};\]initial velocity\[\vec{v}={{v}_{0}}\hat{i}\] Force acting on electron due to electric field \[\vec{F}=(-e(-{{E}_{0}}\hat{i})=e{{E}_{0}}\hat{i}\] Acceleration produced in the electron,\[\vec{a}=\frac{{\vec{F}}}{m}=\frac{e{{E}_{0}}}{m}\hat{i}\] Now, velocity of electron after time t, \[{{\vec{v}}_{t}}=\vec{v}+\vec{a}t=\left( {{v}_{0}}+\frac{e{{E}_{0}}t}{m} \right)\hat{i}\]or\[|{{\vec{v}}_{t}}|={{v}_{0}}+\frac{e{{E}_{0}}t}{m}\] Now,\[{{\lambda }_{t}}=\frac{h}{m{{v}_{t}}}=\frac{h}{m\left( {{v}_{0}}+\frac{e{{E}_{0}}t}{m} \right)}=\frac{h}{m{{v}_{0}}\left( 1+\frac{e{{E}_{0}}t}{m{{v}_{0}}} \right)}\] \[=\frac{{{\lambda }_{0}}}{\left( 1+\frac{e{{E}_{0}}t}{m{{v}_{0}}} \right)}\] \[\left( \because {{\lambda }_{0}}=\frac{h}{m{{v}_{0}}} \right)\]You need to login to perform this action.
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