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(C) The initial de Broglie wavelength is $\lambda_0 = \frac{h}{m_0v_0}$.When the wavelength becomes half,$\lambda = \frac{\lambda_0}{2} = \frac{h}{2m_

Vedclass · Accepted Answer

$\sqrt{3} \frac{m_0v_0}{q_0E_0}$

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(C) The initial de Broglie wavelength is $\lambda_0 = \frac{h}{m_0v_0}$.When the wavelength becomes half,$\lambda = \frac{\lambda_0}{2} = \frac{h}{2m_0v_0}$.Since $\lambda = \frac{h}{p}$,the final momentum $p$ must be $2m_0v_0$.Let the velocity at time $t$ be $v$. Then $mv = 2m_0v_0$,so $v = 2v_0$.The velocity components are $v_y = v_0$ (constant) and $v_x = a_x t = \frac{q_0E_0}{m_0}t$.The magnitude of velocity is $v = \sqrt{v_x^2 + v_y^2}$.Squaring both sides,$v^2 = v_x^2 + v_y^2 \Rightarrow (2v_0)^2 = v_x^2 + v_0^2$.$4v_0^2 = v_x^2 + v_0^2 \Rightarrow v_x^2 = 3v_0^2 \Rightarrow v_x = \sqrt{3}v_0$.Substituting $v_x = \frac{q_0E_0}{m_0}t$,we get $\frac{q_0E_0}{m_0}t = \sqrt{3}v_0$.Therefore,$t = \sqrt{3} \frac{m_0v_0}{q_0E_0}$.

$A$ charged particle $q_0$ of mass $m_0$ is projected along the $y-$axis at $t = 0$ from the origin with a velocity $v_0$. If a uniform electric field $E_0$ also exists along the $x-$axis,then the time at which the de Broglie wavelength of the particle becomes half of its initial value is:

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