Answer to Question #101626 in Atomic and Nuclear Physics for Sridhar

Question

The half life of neutron is 693 seconds.What fraction of neutrons will decay when a beam of neutrons having kinetic energy 0.084 eV travells a distance of 1km? (mass of neutron =1.68times10^(-27)kg and ln2=0.693)
Ans 25×10^-5

Answer 1

$T=693 \; s$

$KE=0.084 \;eV=0.084 \times 1.6\times 10^{-19} \; J$

$s=1000 \;m$

$m=1.68\times 10^{-27} \; kg$

$\ln 2=0.693$

Solution:

$KE=\frac{mv^2}{2}$

$v=\sqrt {\frac{2KE}{m}} =\sqrt {\frac{2\times0.084\times1.6\times 10^{-19}}{1.68\times 10^{-27}}}=4\times 10^3 \; m/s$

Time to travel distance $s=1000 \;m:$

$t=\frac{s}{v}=\frac{1000}{4\times 10^3}=0.25 \;s$

Exponential decay equation:

$N(t)=N_02^{\frac{-t}{T}}$

$t=0.25\; s: \;$

$\frac{N}{N_0}=2^\frac{-0.25}{693}=2^{-3.607\times 10^{-4}}=0.99975$ - fraction of neutrons that won’t decay

$1-\frac{N}{N_0} =1-0.99975=0.00025$ - fraction of neutrons that will decay

Answer: $25\times 10^{-5}$