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(A) The force on the electron is $\vec{F} = q\vec{E} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$.The acceleration of the electron is $a = \frac{F}{m} = \frac

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$\frac{\lambda_0}{[1 + \frac{2E_0 e t}{m v_0}]}$

Vedclass · Answer

(A) The force on the electron is $\vec{F} = q\vec{E} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$.The acceleration of the electron is $a = \frac{F}{m} = \frac{2eE_0}{m}$.The velocity of the electron at time $t$ is given by $v(t) = v_0 + at = v_0 + \left(\frac{2eE_0}{m}\right)t = v_0 \left[1 + \frac{2eE_0 t}{m v_0}\right]$.The de Broglie wavelength at time $t$ is $\lambda = \frac{h}{mv(t)}$.Substituting $v(t)$, we get $\lambda = \frac{h}{m v_0 [1 + \frac{2eE_0 t}{m v_0}]} = \frac{\lambda_0}{[1 + \frac{2eE_0 t}{m v_0}]}$.

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ $(E_0 = \text{constant} > 0)$, with an initial velocity $\vec{V} = v_0\hat{i}$ $(v_0 = \text{constant} > 0)$. If $\lambda_0 = \frac{h}{mv_0}$, its de Broglie wavelength at time $t$ is . . . . . . .

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