A positive,singly ionized atom of mass number | vedclass

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(D) The ion is accelerated through a potential $V$,so its kinetic energy is $K = qV = \frac{1}{2}mv^2$. The momentum $p = \sqrt{2mqV}$.In the magnetic

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$A, B, C$

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(D) The ion is accelerated through a potential $V$,so its kinetic energy is $K = qV = \frac{1}{2}mv^2$. The momentum $p = \sqrt{2mqV}$.In the magnetic field,the ion follows a semi-circular path of radius $R = \frac{p}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$.The ion hits the detector at a distance $x = 2R = \frac{2}{B}\sqrt{\frac{2mV}{q}}$.Given $m = A_M 	imes (5/3) 	imes 10^{-27} 	ext{ kg}$,$V = 192 	ext{ V}$,$q = 1.6 	imes 10^{-19} 	ext{ C}$,$B = 0.1 	ext{ T}$.$(A)$ For $H^{+}$ $(A_M = 1)$: $x = \frac{2}{0.1}\sqrt{\frac{2 	imes (5/3) 	imes 10^{-27} 	imes 192}{1.6 	imes 10^{-19}}} = 20 	imes \sqrt{4 	imes 10^{-8}} = 20 	imes 2 	imes 10^{-4} 	ext{ m} = 4 	imes 10^{-3} 	ext{ m} = 0.4 	ext{ cm}$. Wait,re-calculating: $x = 20 	imes \sqrt{\frac{640}{1.6} 	imes 10^{-8}} = 20 	imes \sqrt{400 	imes 10^{-8}} = 20 	imes 20 	imes 10^{-4} = 0.04 	ext{ m} = 4 	ext{ cm}$. So $(A)$ is correct.$(B)$ For $A_M = 144$: $x = 4 	ext{ cm} 	imes \sqrt{144} = 4 	imes 12 = 48 	ext{ cm}$. So $(B)$ is correct.$(C)$ For $A_M = 1$,$x_0 = 4 	ext{ cm}$. For $A_M = 196$,$x_1 = 4 	ext{ cm} 	imes \sqrt{196} = 4 	imes 14 = 56 	ext{ cm}$. The height is $x_1 - x_0 = 56 - 4 = 52 	ext{ cm}$. So $(C)$ is correct.$(D)$ For $A_M = 196$,$R = x/2 = 56/2 = 28 	ext{ cm}$. So $(D)$ is incorrect.

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