question_answer

An electron of mass m with an initial velocity\[\overrightarrow{v}={{v}_{0}}\widehat{i}({{v}_{0}}>0)\] enters an electric field \[\overrightarrow{\text{E}}\text{=-}{{\text{E}}_{\text{0}}}\widehat{\text{i}}\text{(}{{\text{E}}_{\text{0}}}\text{=}\]constant > 0) at t = 0. If \[{{\lambda }_{0}}\] is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is                     [NEET - 2018]

A)  \[{{\lambda }_{\text{0}}}\text{t}\]                   

B)       \[{{\lambda }_{0}}\left( 1+\frac{e{{E}_{0}}}{m{{V}_{0}}}t \right)\]

C)  \[\frac{{{\lambda }_{0}}}{\left( 1+\frac{e{{E}_{0}}}{m{{V}_{0}}}t \right)}\]   

D)       \[{{\lambda }_{0}}\]

Correct Answer: C

Solution :

[c] Initial de-Broglie wavelength

\[{{\lambda }_{0}}\text{=}\frac{\text{h}}{\text{m}{{\text{V}}_{\text{0}}}}\]                                 (i)

Acceleration of electron

\[a=\frac{e{{E}_{0}}}{m}\]

Velocity after time 't'

\[\text{V=}\left( {{\text{V}}_{\text{0}}}\text{+}\frac{\text{e}{{\text{E}}_{\text{0}}}}{\text{m}}\text{t} \right)\]

So, \[\lambda \text{=}\frac{\text{h}}{\text{mV}}\text{=}\frac{\text{h}}{\text{m}\left( {{\text{V}}_{\text{0}}}\text{+}\frac{\text{e}{{\text{E}}_{\text{0}}}}{\text{m}}\text{t} \right)}\]

\[=\frac{h}{m{{V}_{0}}\left[ 1+\frac{e{{E}_{0}}}{m{{V}_{0}}}t \right]}=\frac{{{\lambda }_{0}}}{\left[ 1+\frac{e{{E}_{0}}}{m{{V}_{0}}}t \right]}\]    (ii)

Divide (ii) by (i),

\[\lambda =\frac{{{\lambda }_{0}}}{\left[ 1+\frac{e{{E}_{0}}}{m{{V}_{0}}}t \right]}\]