A proton with a kinetic energy of 2.0 , eV mo | vedclass

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(C) The kinetic energy $K = 2.0 \, eV = 2.0 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-19} \, J$.The velocity $v$ is given by $K = \frac{1}{2}m

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The kinetic energy $K = 2.0 \, eV = 2.0 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-19} \, J$.The velocity $v$ is given by $K = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 	imes 3.2 	imes 10^{-19}}{1.6 	imes 10^{-27}}} = \sqrt{4 	imes 10^8} = 2 	imes 10^4 \, m/s$.The pitch $p$ of the helical path is given by $p = (v \cos 	heta) 	imes T$,where $T = \frac{2\pi m}{qB}$ is the time period.Substituting the values: $p = v \cos 60^{\circ} 	imes \frac{2\pi m}{qB}$.$p = (2 	imes 10^4) 	imes \frac{1}{2} 	imes \frac{2\pi 	imes 1.6 	imes 10^{-27}}{1.6 	imes 10^{-19} 	imes (\frac{\pi}{2} 	imes 10^{-3})}$.$p = 10^4 	imes \frac{2 	imes 10^{-27}}{10^{-19} 	imes 0.5 	imes 10^{-3}} = 10^4 	imes 4 	imes 10^{-5} = 0.4 \, m$.Converting to centimeters: $p = 0.4 	imes 100 = 40 \, cm$.

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