The half-life of a particle of mass 1.6 times | vedclass

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(D) The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.Given $K = 0.05 \,eV = 0.05 	imes 1.6 	imes 10^{-19} \,J$ and $m = 1.6

Vedclass · Accepted Answer

$0.0001$

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(D) The kinetic energy $K$ of the particle is given by $K = \frac{1}{2}mv^2$.Given $K = 0.05 \,eV = 0.05 	imes 1.6 	imes 10^{-19} \,J$ and $m = 1.6 	imes 10^{-26} \,kg$.$0.05 	imes 1.6 	imes 10^{-19} = \frac{1}{2} 	imes 1.6 	imes 10^{-26} 	imes v^2$.$v^2 = \frac{0.05 	imes 1.6 	imes 10^{-19} 	imes 2}{1.6 	imes 10^{-26}} = 0.1 	imes 10^7 = 10^6 \,m^2/s^2$.$v = 10^3 \,m/s$.The time $t$ taken to travel a distance $D = 1 \,m$ is $t = \frac{D}{v} = \frac{1}{10^3} = 10^{-3} \,s$.The number of half-lives $n$ is $n = \frac{t}{T_{1/2}} = \frac{10^{-3}}{6.9} \approx 1.45 	imes 10^{-4}$.The fraction of particles remaining is $N/N_0 = (1/2)^n = 2^{-n}$.The fraction decayed is $1 - 2^{-n} = 1 - e^{-n \ln 2} \approx 1 - (1 - n \ln 2) = n \ln 2$.Since $n \approx 1.45 	imes 10^{-4}$ and $\ln 2 \approx 0.69$,the fraction decayed $\approx 1.45 	imes 10^{-4} 	imes 0.69 \approx 10^{-4} = 0.0001$.

The half-life of a particle of mass $1.6 \times 10^{-26} \,kg$ is $6.9 \,s$. $A$ stream of such particles is travelling with a kinetic energy of $0.05 \,eV$ per particle. The fraction of particles that will decay when they travel a distance of $1 \,m$ is

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