The charge on a particle Y is double the char | vedclass

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(C) The radius of a circular path of a charged particle accelerated through a potential difference $V$ in a magnetic field $B$ is given by $r = \frac{

Vedclass · Accepted Answer

$\frac{R_1^2}{2R_2^2}$

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(C) The radius of a circular path of a charged particle accelerated through a potential difference $V$ in a magnetic field $B$ is given by $r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$.Since $B$ and $V$ are the same for both particles,we have $r \propto \sqrt{\frac{m}{q}}$.Therefore,$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y} \cdot \frac{q_Y}{q_X}}$.Given that $q_Y = 2q_X$,we substitute this into the equation:$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y} \cdot \frac{2q_X}{q_X}} = \sqrt{\frac{m_X}{m_Y} \cdot 2}$.Squaring both sides,we get $\frac{R_1^2}{R_2^2} = \frac{m_X}{m_Y} \cdot 2$.Thus,the ratio of the mass of $X$ to that of $Y$ is $\frac{m_X}{m_Y} = \frac{R_1^2}{2R_2^2}$.

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