Question

A $2\, MeV$ neutron is emitted in a fission reactor. If it looses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of $0.039\, eV$

• A. 20
• B. 26
• C. 30
• D. 42
• E. 48

Answer 1

Answer: (B)

$2\, MeV$ neutron looses half of its kinetic energy in each collision. So, making geometric progression of each collision
$3\, MeV ,\ ,l MeV ,\, 5\, MeV\, ...0.039\, eV$
In this geometric progression,
$a=2\, MeV$
$\gamma=\frac{1}{2}$
$n$ th term,
$a_{n}=a r^{n-1}$
$\therefore 0.039=2 \times 10^{6}(1 / 2)^{n-1}$
$\Rightarrow 1.95 \times 10^{-8}=(1 / 2)^{n-1}$
Taking log both sides,
$\Rightarrow (n-1) \ln \left(\frac{1}{2}\right)=\ln \left(1.95 \times 10^{-8}\right)$
Number of collision $(n)=26.6 \approx 26$ collision