An electron (mass m) is moving with initial velocity v→=v0i^+v0j^ in an electric field E→=−E0k^ . If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by

λ=hm2V0λ1=hV02+V02+(eE0tm)2⇒λ1=λ01+e2E02t22m2V02

An electron (mass m) is moving with initial velocity v→=v0i^+v0j^ in an electric field E→=−E0k^ . If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by

λ=hm2V0λ1=hV02+V02+(eE0tm)2⇒λ1=λ01+e2E02t22m2V02

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An electron (mass m) is moving with initial velocity $\vec{v} = v_{0} \hat{i} + v_{0} \hat{j}$ in an electric field $\vec{E} = - E_{0} \hat{k}$ . If $λ_{0}$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by

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

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

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

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

answer is C.

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$λ = \frac{h}{m \sqrt{2} V_{0}}$

$λ^{1} = \frac{h}{\sqrt{V_{0}^{2} + V_{0}^{2} + {(\frac{e E_{0} t}{m})}^{2}}}$

$\Rightarrow λ^{1} = \frac{λ_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{2 m^{2} V_{0}^{2}}}}$

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An electron (mass m) is moving with initial velocity v→=v0i^+v0j^ in an electric field E→=−E0k^ . If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by

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An electron (mass m) is moving with initial velocity $\vec{v} = v_{0} \hat{i} + v_{0} \hat{j}$ in an electric field $\vec{E} = - E_{0} \hat{k}$ . If $λ_{0}$ is initial de-Broglie wavelength of electron, its de-Broglie wavelength at time t is given by