Three ions H^+,He^+,and O^{2+} having the sam | vedclass

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

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$He^+$ and $O^{2+}$ will be deflected equally.

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(B) The radius of the circular path of a charged particle in a uniform 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: $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.For constant $K$ and $B$,the radius $r \propto \frac{\sqrt{m}}{q}$.Deflection is inversely proportional to the radius $(d \propto \frac{1}{r})$,so $d \propto \frac{q}{\sqrt{m}}$.For $H^+$: $q=1, m=1 \Rightarrow d \propto \frac{1}{\sqrt{1}} = 1$.For $He^+$: $q=1, m=4 \Rightarrow d \propto \frac{1}{\sqrt{4}} = 0.5$.For $O^{2+}$: $q=2, m=16 \Rightarrow d \propto \frac{2}{\sqrt{16}} = \frac{2}{4} = 0.5$.Comparing the values,$H^+$ has the highest deflection,while $He^+$ and $O^{2+}$ have equal and lower deflection.

Three ions $H^+$,$He^+$,and $O^{2+}$ having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. Then:

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