Question

The nuclear reaction $n{+}_5^{10}B{\to }_3^{7}Li{+}_2^{4}He$ is observed to occur even when very slow-moving neutrons $(M_n=1.0087amu)$ strike a boron atom at rest. For a particular reaction in which $K_n=0$, the helium $(M_{He}=4.0026amu)$ is observed to have a speed of $9.30\times {10}^{6}m{s}^{-1}$. Determine (a) the kinetic energy of the lithium $(M_{Li}=7.0160amu)$ and (b) the Q value of the reaction.

Answer 1

Energy released by the reaction $=Q=△m{c}^2$

$Q=△m=1.0087+10.811-7.0160-4.0026=0.8011Q=△m{c}^2=0.8011\times 931=745.82MeV=754.82\times 1.6\times {10}^{-13}=1193.3\times {10}^{-13}J$

K.E. of lithium $=\frac{1}{2}m{v}^2=\frac{1}{2}\times 7.0160\times (9.30\times {10}^6{)}^2=303.40\times {10}^{12}J$