Two ions of masses 4 , amu and 16 , amu have | vedclass

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(B) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the

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lighter ion will be deflected more than heavier ion

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(B) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.Since $K$ and $B$ are constant for both ions,$R \propto \frac{\sqrt{m}}{q}$.For the lighter ion $(m_1 = 4, q_1 = 2)$: $R_1 \propto \frac{\sqrt{4}}{2} = 1$.For the heavier ion $(m_2 = 16, q_2 = 3)$: $R_2 \propto \frac{\sqrt{16}}{3} = \frac{4}{3}$.Since $R_2 > R_1$,the radius of the path of the heavier ion is larger.The deflection $	heta$ is related to the path radius $R$ by $\sin 	heta = \frac{d}{R}$,where $d$ is the width of the magnetic field region.Since $	heta \propto \frac{1}{R}$,a smaller radius $R$ corresponds to a larger deflection $	heta$.Therefore,the lighter ion (with smaller $R$) will be deflected more than the heavier ion.

Two ions of masses $4 \, amu$ and $16 \, amu$ have charges $+2e$ and $+3e$ respectively. These ions pass through a region of constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then:

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