Question

A proton $X$ is projected in a region of uniform electric field $E$ with speed $u$. Another proton $Y$ is projected simultaneously with same speed $u$ in a separate region having a uniform magnetic field $B$. It was found that both $X$ and $Y$ had same de-Broglie wavelength after time $T$. Specific charge on a proton is $\sigma \text{(C/kg)}$.

• A. The angle $\alpha$ at which the proton $X$ was projected with respect to the direction of the electric field is $\frac{\pi }{2}+{\sin }^{-1}\left[\frac{eEt}{mu}\right]$
• B. The angle $\alpha$ at which the proton $X$ was projected with respect to the direction of the electric field is $\frac{\pi }{2}+{\sin }^{-1}\left[\frac{eEt}{2mu}\right]$
• C. The displacement of the proton $X$ in time $T$ is $\sqrt{u^2-\frac{{\sigma }^2{E}^2{T}^2}{8}}.T$
• D. The displacement of the proton $X$ in time $T$ is $\sqrt{u^2-\frac{{\sigma }^2{E}^2{T}^2}{4}}.T$

Answer 1

Answer: (B), (D)

The correct options are
B The angle $\alpha$ at which the proton $X$ was projected with respect to the direction of the electric field is $\frac{\pi }{2}+{\sin }^{-1}\left[\frac{eEt}{2mu}\right]$
D The displacement of the proton $X$ in time $T$ is $\sqrt{u^2-\frac{{\sigma }^2{E}^2{T}^2}{4}}.T$

Speed of $Y$ does not change as it is projected in magnetic field.

De-Broglie wavelength of $X$ will become equal to that of $Y$ when speed of $X$ again becomes equal to its initial speed $u$. This is possible in a situation shown in figure. Proton starts from $A$ with speed $u$ reaches point $B$ with same speed $u$.
$T=$ time of flight from $A$ to $B$

$T=\frac{2u_y}{a}$ $[a=\frac{eE}{m}]$
$\therefore u_y=\frac{eET}{2m}$
(or)
$u\sin \theta =\frac{eET}{2m}$
$\theta ={\sin }^{-1}\left[\frac{eET}{2mu}\right]$
Desired angle between angle of projection and electric field is$\alpha =\frac{\pi }{2}+\theta$
$\alpha =\frac{\pi }{2}+{\sin }^{-1}\left[\frac{eET}{2mu}\right]$


The displacement of the proton $X$ in time $T$ is
$AB=u_{x}T=\sqrt{u^2-u_y^2}.T(\because u_x=\sqrt{u^2-u_y^2})AB=\sqrt{u^2-\frac{{\sigma }^2{E}^2{T}^2}{4}}.T$
$\Rightarrow AB=\sqrt{u^2-\frac{{\sigma }^2{E}^2{T}^2}{4}}.T$