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An electron (mass m ) with initial velocity \vec{v} = v_{0}\widehat{i} + v_{0}\widehat{j} is in an electric field \vec{E} = -E_{0}\widehat{k}. If \lambda_{0} is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by :
Option: 1 \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}

Option: 2 \frac{\lambda_{0}}{\sqrt{2 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}

Option: 3 \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{2m^{2}v_{0}^{2}}}}

Option: 4 \frac{\lambda_{0} \sqrt{2}}{\sqrt{1 + \frac{e^{2} E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}
 

Answers (1)

best_answer

 

 

  

\text { Initially } m\left(\sqrt{2} v_{0}\right)=\frac{h}{\lambda_{0}}

 

We know v=u+at and here a=\frac{qE}{m}=\frac{eE_0}{m}

 

\text { Velocity as a function of time }=v_{0} \hat{i}+v_{0} \hat{j}+\frac{e E_{0}}{m} t \hat{k}

 

So, magnitude of velocity:- \sqrt{2 v_{0}^{2}+\frac{e^{2} E_{0}^{2}}{m^{2}} t^{2}}$

 

\text { so wavelength } \lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2 \mathrm{v}_{0}^{2}+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2}}{\mathrm{m}^{2}} \mathrm{t}^{2}}}

 

\Rightarrow \lambda=\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2}}{2\mathrm{m}^{2} \mathrm{v}_{0}^{2}} \mathrm{t}^{2}}}$

Hence the correct option is (3).

Posted by

vishal kumar

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