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(C) The speed of the charged particle just before entering the magnetic field is $v_0$.By the work-energy theorem,the kinetic energy gained is equal t

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$\frac{qB^2d^2}{8V}$

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(C) The speed of the charged particle just before entering the magnetic field is $v_0$.By the work-energy theorem,the kinetic energy gained is equal to the work done by the electric field:$qV = \frac{1}{2}mv_0^2$Solving for $v_0$:$v_0 = \sqrt{\frac{2qV}{m}}$The radius $r$ of the circular path in the magnetic field is given by:$r = \frac{mv_0}{qB}$From the figure,the distance $d$ between the holes is the diameter of the circular path,so $r = \frac{d}{2}$.Substituting this into the radius formula:$\frac{d}{2} = \frac{m}{qB} \sqrt{\frac{2qV}{m}}$Squaring both sides:$\frac{d^2}{4} = \frac{m^2}{q^2B^2} \cdot \frac{2qV}{m}$Simplifying the expression:$\frac{d^2}{4} = \frac{2mV}{qB^2}$Solving for $m$:$m = \frac{qB^2d^2}{8V}$

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In a Thomson mass spectrograph,the electric field and magnetic field are applied:

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