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An electron (mass m) with an initial velocity $v=v_{0}\hat{i}$ is in an electric field $E=E_{0}\hat{i}$ IF $\lambda_{0}=\frac{h}{mv_{0}}$ its de-broglie wavelength at time t is given by

$(a)\lambda_{0}\\(b)\lambda_{0}\sqrt{1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}}}\\(c)\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}}}}\\ (d)\frac{\lambda_{0}}{\left (1+\frac{e^{2}E_{0}^{2}t^{0}}{m^{2}v_{0}^{2}} \right )}$

Answers (1)

The correct answer is option (c)

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

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infoexpert24

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