JEE Main On 08 April 2018 Question 1
Question: It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $ {P_d} $ ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $ {P_c} $ . The values of $ {P_d} $ and $ {P_c} $ are respectively: [JEE Main Online 08-04-2018]
Options:
A) $ \text{(0, 0)} $
B) $ \text{(0, 1)} $
C) $ \text{(}\cdot 89\text{, }\cdot \text{28)} $
D) $ \text{(}\cdot 28\text{, }\cdot \text{89)} $
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Answer:
Correct Answer: C
Solution:
Conservation of momentum $ mv _0+0=mv _1+2mv _2 $ $ v _0=(v _1+2v _2)…..(i) $
$ e=\frac{v _2-v _1}{v _0}(e=1) $ $ v _2-v _1=v _0…(ii) $
$ v _0-2v _1=2v _0+v _1 $ $ 3v _1=-v _0\Rightarrow v _1=\frac{-v _0}{3} $
Fractional loss of its K.E. $ =\frac{\frac{1}{2}mv _0^{2}-\frac{1}{2}m{{( \frac{v _0}{3} )}^{2}}}{\frac{1}{2}mv _0^{2}}=\frac{8}{9}=0.88\approx 0.89 $
Neutron colliding with carbon Conservation of momentum $ mv _0=mv _1+12mv _2 $
$ v _1+12v _2=v _0…(i) $ $ e=1=\frac{V _2-V _1}{V _0} $ $ V _2-V _1=V _0…(ii) $
$ \Rightarrow V _1+12v _1=v _0-12v _0 $ $ =\frac{-11v _0}{13} $
Fractional loss in K.E. $ =\frac{\frac{1}{2}mv _0^{2}-\frac{1}{2}m{{( \frac{11}{13}v _0 )}^{2}}}{\frac{1}{2}mv _0^{2}}=1-\frac{121}{169}=\frac{48}{169} $ $ \approx 0.28 $
