A proton and an alpha-particle having equal k | vedclass

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(B) The force on a charged particle in an electric field is given by $F = qE$,where $q$ is the charge and $E$ is the electric field intensity.The acce

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$\alpha$-particle trajectory is more curved.

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(B) The force on a charged particle in an electric field is given by $F = qE$,where $q$ is the charge and $E$ is the electric field intensity.The acceleration of the particle is $a = \frac{F}{m} = \frac{qE}{m}$.For a particle moving with initial velocity $v$ perpendicular to the electric field,the deflection $y$ after traveling a horizontal distance $x$ is given by $y = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{x}{v})^2 = \frac{qE x^2}{2 m v^2}$.Since the kinetic energy $K = \frac{1}{2} m v^2$ is constant,we can write $m v^2 = 2K$. Substituting this into the deflection equation,we get $y = \frac{qE x^2}{4K}$.For a proton,$q_p = e$. For an $\alpha$-particle,$q_{\alpha} = 2e$. Since both have the same kinetic energy $K$ and are in the same electric field $E$,the deflection $y$ is directly proportional to the charge $q$.Since $q_{\alpha} > q_p$,the deflection of the $\alpha$-particle is greater than that of the proton. Therefore,the $\alpha$-particle trajectory is more curved.

$A$ proton and an $\alpha$-particle having equal kinetic energy are projected into a uniform transverse electric field as shown in the figure.

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