A mixed beam of He^+ and O^{2+} ions (mass of | vedclass

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(C) 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^

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All the ions will be deflected equally

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(C) 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 this into the radius formula,we get $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.Given that $K$ and $B$ are constant for all ions,the radius $r \propto \frac{\sqrt{m}}{q}$.For $He^+$ ions: $m_1 = 4 \ amu$,$q_1 = 1e$.For $O^{2+}$ ions: $m_2 = 16 \ amu$,$q_2 = 2e$.Calculating the ratio of radii: $\frac{r_{He^+}}{r_{O^{2+}}} = \sqrt{\frac{m_1}{m_2}} 	imes \frac{q_2}{q_1} = \sqrt{\frac{4}{16}} 	imes \frac{2}{1} = \frac{1}{2} 	imes 2 = 1$.Since the radii of the paths are equal,all ions will be deflected equally.

$A$ mixed beam of $He^+$ and $O^{2+}$ ions (mass of $He^+ = 4 \ amu$ and that of $O^{2+} = 16 \ amu$) passes through a region of a constant perpendicular magnetic field. If the kinetic energy of all the ions is the same,then:

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