Question

A particle of charge q and mass m enters a region of a transverse electric field of $E_0\hat{j}$ with initial velocity $v_o\hat{i}$. 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.

• A. $\frac{q}{m}$
• B. $\frac{m}{q}$
• C. $\sqrt{\frac{q}{m}}$
• D. $\sqrt{\frac{m}{q}}$

Answer 1

The correct option is B $\frac{m}{q}$

$F=ma=qE$

$m\ddot{x}=0$

$x=vot$ ....(1)

$m\ddot{y}=q{E}_0$

$\Rightarrow \ddot{y}=\frac{q{E}_0}{m}$

$y=\frac{qeot}{m}$ ...(2)

$y=\frac{qEo{t}^2}{m}$

velocity $v=\sqrt{v{x}^2+v{y}^2}=\sqrt{(Vo{)}^2+{\left(\frac{qEot}{m}\right)}^2}$

$K.E=\frac{1}{2}m{v}^2=\frac{1}{2}m(v{o}^2+{\left(\frac{qeot}{m}\right)}^2)$

$\Delta (K.E.)=\frac{1}{2}m{\left(\frac{q{E}_o\Delta t}{m}\right)}^2$

$=\frac{qE{o}^2(\Delta t{)}^2}{2m}$

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m{E}_k}}$.

$\lambda -0\to \frac{{\lambda }_0}{3}$

$E_k\to q{E}_n$. $\Delta E_k=8E_k$

At $t=0$ $K.E=\frac{1}{2}mv{o}^2=E_{k_1}$

$t=t$ $K.E=q{E}_{k_1}$

$\Delta K.E=8E_{k_1}=\frac{8}{2}mv{o}^2=4mv{o}^2$

$mv{o}^2=\frac{1}{2}m{\left(\frac{q{E}_o\Delta t}{m}\right)}^2$

$\Rightarrow v{o}^2=\frac{1}{2m_2}{q}^2{E}_o^2\delta t^2$

$\to v{o}^2=\frac{q^2{E}_o^2\Delta t^2}{2m_2}$ $\Delta t=t$

$v{o}^2=\frac{q^2{E}_o^2\Delta t^2}{2m_2}$

$\Delta t^2=C\frac{2m^2}{q^2{E}_o^2}$

$\Delta t\propto \left(\frac{m}{q}\right)$