{H^ + },,H{e^ + } and {O^{ + + }} ions having | vedclass

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(D) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

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(D) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}}$.Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.Since $K$ and $B$ are constant,$r \propto \frac{\sqrt{m}}{q}$.Given mass ratios $m_H:m_{He}:m_O = 1:4:16$ and charges $q_H=1, q_{He}=1, q_O=2$ (in units of $e$):$r_H \propto \frac{\sqrt{1}}{1} = 1$$r_{He} \propto \frac{\sqrt{4}}{1} = 2$$r_O \propto \frac{\sqrt{16}}{2} = \frac{4}{2} = 2$Thus,the ratio of radii is $r_H:r_{He}:r_O = 1:2:2$.Since deflection is inversely proportional to the radius $(d \propto 1/r)$,the ion with the smallest radius $(H^+)$ is deflected the most,and $He^+$ and $O^{++}$ have the same radius,hence the same deflection. Therefore,both $(a)$ and $(c)$ are correct.

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