question_answer

A neutron with velocity V strikes a stationary deuterium atom, its kinetic energy changes by a factor of [DCE 2000]

A)            \[\frac{15}{16}\]                 

B)            \[\frac{1}{2}\]

C)            \[\frac{2}{1}\]                      

D)            None of these                     

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  \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}(2m)v_{2}^{2}\]                           ... (ii) By solving (i) and (ii) we get \[{{v}_{1}}=\frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}v+\frac{2{{m}_{2}}}{({{m}_{1}}+{{m}_{2}})}v\] Þ \[{{v}_{1}}=\frac{{{m}_{1}}+2m}{3m}\]\[=-\frac{v}{3}\] \[{{K}_{i}}=\frac{1}{2}m{{v}^{2}}\],  \[{{K}_{f}}=\frac{1}{2}mv_{1}^{2}\] \[\Rightarrow \frac{{{K}_{i}}-{{K}_{f}}}{{{K}_{i}}}=1-\frac{v_{1}^{2}}{{{v}^{2}}}\] \[=1-\frac{1}{9}=\frac{8}{9}\] (Fractional change in K.E.)