Modern Physics Ques 142
- There is a stream of neutrons with a kinetic energy of $0.0327 eV$. If the half-life of neutrons is $700 s$, what fraction of neutrons will decay before they travel a distance of 10 $m$ ?
$(1986,6$ M)
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Answer:
Correct Answer: 142.$3.96 \times 10^{-6}$
Solution:
Formula:
- Speed of neutrons $=\sqrt{\frac{2 K}{m}} \quad$ From $K=\frac{1}{2} m v^{2}$
$$ \text { or } \quad v=\sqrt{\frac{2 \times 0.0327 \times 1.6 \times 10^{-19}}{1.675 \times 10^{-27}}} $$
$$ \approx 2.5 \times 10^{3} m / s $$
Time taken by the neutrons to travel a distance of $10 m$ :
$$ t=\frac{d}{v}=\frac{10}{2.5 \times 10^{3}}=4.0 \times 10^{-3} $$
Number of neutrons decayed after time
$$ t: N=N _0\left(1-e^{-\lambda t}\right) $$
$\therefore$ Fraction of neutrons that will decay in this time interval
$$ \begin{aligned} & =\frac{N}{N _0}=\left(1-e^{-\lambda t}\right)=1-e^{-\frac{\ln (2)}{700} \times 4.0 \times 10^{-3}} \\ & =3.96 \times 10^{-6} \end{aligned} $$
BETA
