The half-life of a stream of radioactive part | vedclass

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(A) Given,$K.E. = 4 	ext{ eV} = 4 	imes 1.6 	imes 10^{-19} 	ext{ J} = 6.4 	imes 10^{-19} 	ext{ J}$.Mass $m = 3.2 	imes 10^{-21} 	ext{ kg}$.Usi

Vedclass · Accepted Answer

$87.5$

Vedclass · Answer

(A) Given,$K.E. = 4 	ext{ eV} = 4 	imes 1.6 	imes 10^{-19} 	ext{ J} = 6.4 	imes 10^{-19} 	ext{ J}$.Mass $m = 3.2 	imes 10^{-21} 	ext{ kg}$.Using $K.E. = \frac{1}{2}mv^2$,we get $v = \sqrt{\frac{2 	imes K.E.}{m}} = \sqrt{\frac{2 	imes 6.4 	imes 10^{-19}}{3.2 	imes 10^{-21}}} = \sqrt{4 	imes 10^2} = 20 	ext{ m/s}$.Time taken to travel $D = 3.6 	ext{ km} = 3600 	ext{ m}$ is $t = \frac{D}{v} = \frac{3600}{20} = 180 	ext{ s} = 3 	ext{ minutes}$.Since the half-life $T_{1/2} = 1 	ext{ minute}$,the time $t = 3 	ext{ half-lives}$.The number of particles remaining is $N = N_0 \left(\frac{1}{2}ight)^n = N_0 \left(\frac{1}{2}ight)^3 = \frac{N_0}{8}$.The number of particles decayed is $N_0 - N = N_0 - \frac{N_0}{8} = \frac{7}{8}N_0$.The percentage of decayed particles is $\frac{7/8 N_0}{N_0} 	imes 100 = 87.5\%$.

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