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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The nuclear reaction $\;n + _{5}^{10}B \to _{3}^{7}{Li}+_{2}^{4}He\;$ is observed to occur when very slow - moving neutrons $\;(M_{n}=1.0087 u)\;$ strike a boron atom at rest . For a particular reaction in which $\;K_{n} \approx 0\;,$ the helium $\;(M_{He}=4.0026 u)\;$ is observed to have a speed of $\;9.30\times 10^{6} m/s\;.$ Then the kinetic energy of $\;(lithium\;M_{li}=7.016 u)\;$. Find Q value of reaction .

$(a)\;1.42 MeV\qquad(b)\;3 MeV\qquad(c)\;2 MeV\qquad(d)\;2.82 MeV$

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Answer : (d) $\;2.82 MeV$
Explanation :
We are given $\; K_{\alpha}=K_{\lambda}=0$
So $\;Q=K_{Li}+K_{He}$
Where $\;K_{He}=\large\frac{1}{2}\;M_{He}v_{He}^{2}=\large\frac{1}{2}\;(4.0026 u)\;(1.66\times10^{-27} Kg/u)\times(9.30\times10^{6} m/s)^2$
$=2.87 \times 10^{-13} J =1.8 MeV$
Hence $\;Q=1.02 MeV + 1.8 MeV = 2.82 MeV\;.$
answered Feb 27, 2014 by yamini.v
 

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