An alpha particles kinetic energy `T= 5.3 MeV` initiates a nuclear reaction `Be^(9)(alpha,n)C^(12)` with energy yield `Q=+5.7MeV`. Find the kinetic energy of the nerutron outgoing at right angle to the motion direction of the alpha-particles.

Explanation

It is understood that `Be^(9)` is initially at rest. The moment of the outgoing neturon is `sqrt(2m_(n),T_(n))hat(j)`. The moment of `C^(12)` is
`sqrt(2m_(alpha)T)hat(i)-sqrt(2m_(n)T_(n))hat(j)`
Then by energy conservation
`T+Q=T_(n)+(2m_(alpha)T+2m_(n)T_(n))/(2m_(ci))`
`(m_(c )` is the mass of `C^(12)`)
Thus `T_(n)=(m_(c )(T+Q)-m_(alpha)T)/(m_(c )+m_(n))`
`((m_(c)-m_(alpha))T+m_(c )Q)/(m_(c )+m_(n))=(Q+(1-(m_(alpha))/(m_(c )))T)/(1+(m_(n))/(m_(c )))=8.52MeV`