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  •  An electron (mass ) with an initial velocity <img alt="\mathrm{v}=\mathrm{v}_0 \hat{\mathrm{i}}

 An electron (mass \mathrm{m} ) with an initial velocity \mathrm{v}=\mathrm{v}_0 \hat{\mathrm{i}} is in an electric field  \mathrm{E}=\mathrm{E}_0 \hat{\mathrm{j}}. If  \lambda_0=\mathrm{h} / \mathrm{mv}_0, its deBroglie wavelength at time t is given by:
 

Option: 1

\lambda_0


Option: 2

\quad \lambda_0 \sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0^2}}


Option: 3

\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0^2}}}


Option: 4

\frac{\lambda_0}{\left(1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0^2}\right)}


Answers (1)

best_answer

Initial de-Broglie wavelength of electron, \lambda_0=\frac{\mathrm{h}}{\mathrm{mv}_0}

Force on electron in electric field, F=-\mathrm{eE}=-\mathrm{eE}_0 \hat{\mathrm{j}}

Acceleration of electron, a=\frac{F}{m}=\frac{e E_0 \hat{j}}{m}

It is acting along negative Y-axis.

The initial velocity of electron along X-axis, v_{x_0}=v_0 \hat{i}. Initial velocity of electron Y-axis, v_{y_0}=0

Velocity of electron after time t along X-axis, v_x=v_0 \hat{i} (\because there is not electron along X-axis)

Velocity of electron after time t along Y-axis, v_y=0+\left(-\frac{e E_0}{m} \hat{j}\right) t=-\frac{e E_0}{m} \hat{t}

Magnitude of velocity of electron after time t is v=\sqrt{v_x^2+v_y^2}=\sqrt{v_0^2+\left(\frac{-e E_0}{m} t\right)^2}
\Rightarrow=v_0 \sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0^2}}

de-Broglie wavelength, \lambda^{\prime}=\frac{h}{\mathrm{mv}}

\Rightarrow=\frac{h}{m v_0 \sqrt{1+\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2 /\left(\mathrm{m}^2 \mathrm{v}_0^2\right)}}=\frac{\lambda_0}{\sqrt{1+\mathrm{e}^2 \mathrm{E}_0^2 \mathrm{t}^2 / \mathrm{m}^2 \mathrm{v}_0^2}}

Posted by

Irshad Anwar

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