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  •  An electron (mass m) with an initial velocity <img alt="\mathrm{\overrightarrow{v}=v_{0} \hat{i}\left(v_{0}>0\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

 An electron (mass m) with an initial velocity \mathrm{\overrightarrow{v}=v_{0} \hat{i}\left(v_{0}>0\right)}  is moving in an electric field \mathrm{\overrightarrow{E}=-E_{0} \hat{i}\left(E_{0}>0\right)} where \mathrm{\mathrm{E}_{0}} is constant. If at \mathrm{t=0} de Broglie wavelength is \mathrm{\lambda_{0}=\frac{h}{m v_{0}}},  then its de Broglie wavelength after time t is given by 

Option: 1

\lambda_{0}


Option: 2

\mathrm{\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)}


Option: 3

\mathrm{\lambda_{0} t}


Option: 4

\mathrm{\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}}


Answers (1)

best_answer

\mathrm{\bar{v} =\bar{u}+\bar{a} t }

\mathrm{\bar{v} =v_0 \hat{i}+\frac{(-e)\left(-E_0\right)\hat{i}}{m} t}

\mathrm{\bar{v} =\left(v_0+\frac{e E_0 t}{m}\right) \hat{i} \rightarrow(1) }

\mathrm{\lambda_0 =\frac{h}{m v_0} \quad(\text { Given }) }
At time t,

\mathrm{\lambda =\frac{h}{m v} }

\mathrm{=\frac{h}{m\left(v_0+\frac{e E_0 t}{m}\right)} }

\mathrm{\lambda =\frac{\lambda_0 v_0}{\left(v_0+\frac{e E_0 t}{m}\right)} }

\mathrm{\lambda =\frac{\lambda_0}{\left(1+\frac{e E_0 t}{m v_0}\right)}}

Hence 4 is correct option.




 

Posted by

Ritika Harsh

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