Question

The kinetic energy of a neutron beam is 0.0837 eV. The half-life of neutrons is 693s and the mass of neutrons is $1.675\times {10}^{-27}$ kg. The fraction of decay in travelling a distance of 40m will be

• A. ${10}^{-3}$
  
• B. ${10}^{-4}$
  
• C. ${10}^{-5}$
  
• D. ${10}^{-6}$

Answer 1

The correct option is C ${10}^{-5}$


$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2\times 0.0837\times 1.6\times {10}^{-19}}{1.675\times {10}^{-27}}}=4\times {10}^{3}m/sec$
$\therefore$ Time taken by neutrons to travel a distance of 40 m $\Delta t^{\prime }=\frac{40}{4\times {10}^3}={10}^{-2}sec$
$\therefore$ $\frac{dN}{dt}=\lambda N\Rightarrow \frac{dN}{N}=\lambda t$
$\therefore$ Fraction of neutrons decayed in $\Delta t$ sec in $\frac{\Delta N}{N}=\lambda \Delta t=\frac{0.693}{T}\Delta t=\frac{0.693}{693}\times {10}^{-2}={10}^{-5}$