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

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(B) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{\sqrt{2 m E_k}}{q B}$.Since the kinetic energy $E

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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 $r = \frac{\sqrt{2 m E_k}}{q B}$.Since the kinetic energy $E_k$ and magnetic field $B$ are the same for both ions,the radius $r \propto \frac{\sqrt{m}}{q}$.The deflection of the ion is inversely proportional to the radius of its path,so $	ext{deflection} \propto \frac{1}{r} \propto \frac{q}{\sqrt{m}}$.For the lighter ion $(m_1 = 4 \ amu, q_1 = 2e)$: $	ext{deflection}_1 \propto \frac{2}{\sqrt{4}} = \frac{2}{2} = 1$.For the heavier ion $(m_2 = 16 \ amu, q_2 = 3e)$: $	ext{deflection}_2 \propto \frac{3}{\sqrt{16}} = \frac{3}{4} = 0.75$.Since $1 > 0.75$,the lighter ion will be deflected more than the heavier ion.

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

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