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(B) The initial de-Broglie wavelength is $\lambda_{0} = h / (m v_{0})$.When the wavelength becomes $\lambda_{0} / 3$,the final momentum $p_f$ must be

Vedclass · Accepted Answer

$m/q$

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(B) The initial de-Broglie wavelength is $\lambda_{0} = h / (m v_{0})$.When the wavelength becomes $\lambda_{0} / 3$,the final momentum $p_f$ must be $3 p_0$,which implies the final speed $v_f = 3 v_{0}$.The particle experiences a constant acceleration $a = (q E_{0} / m) \hat{j}$ in the $y$-direction.The velocity at time $t$ is given by $\vec{v} = v_{0} \hat{i} + (q E_{0} t / m) \hat{j}$.The magnitude of the velocity is $v = \sqrt{v_{0}^{2} + (q E_{0} t / m)^{2}}$.Setting $v = 3 v_{0}$,we get $9 v_{0}^{2} = v_{0}^{2} + (q E_{0} t / m)^{2}$.This simplifies to $8 v_{0}^{2} = (q E_{0} t / m)^{2}$.Solving for $t$,we get $t = \frac{2 \sqrt{2} v_{0}}{E_{0}} \cdot \frac{m}{q}$.Therefore,$t \propto m/q$.

$A$ particle of charge $q$ and mass $m$ enters a region of a transverse electric field of $E_{0} \hat{j}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de-Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to

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