Nuclear Physics And Radioactivity Question 39
Question: A neutron with velocity V strikes a stationary deuterium atom, its kinetic energy changes by a factor of [DCE 2000]
Options:
A) $ \frac{15}{16} $
B) $ \frac{1}{2} $
C) $ \frac{2}{1} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Neutron velocity = v, mass = m Deuteron contains 1 neutron and 1 proton, mass = 2m In elastic collision both momentum and K.E. are conserved pi = pf mv = m1v2 + m2v2
Þ mv = mv1 + 2mv2 … (i) By conservation of kinetic energy $ \frac{1}{2}mv^{2}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}(2m)v_{2}^{2} $ … (ii) By solving (i) and (ii) we get $ v_{1}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v+\frac{2m_{2}}{(m_{1}+m_{2})}v $
Þ $ v_{1}=\frac{m_{1}+2m}{3m} $ $ =-\frac{v}{3} $ $ K_{i}=\frac{1}{2}mv^{2} $ , $ K_{f}=\frac{1}{2}mv_{1}^{2} $
$ \Rightarrow \frac{K_{i}-K_{f}}{K_{i}}=1-\frac{v_{1}^{2}}{v^{2}} $ $ =1-\frac{1}{9}=\frac{8}{9} $ (Fractional change in K.E.)
BETA
