In a nuclear reaction, `alpha+"_(7)N^(14)toX+p`
When nitrogen atoms are bombarded by `7.7 Me V alpha-` particles, protons are ejected with a kinetic energy of `5.5 MeV`. The `Q-` value of the reaction is `-1.26 MeV`
(`a`) Find the atomic mass of `X` in amu, identify the element `X`.
(`b`) Find the angle `phi` between the direction of motion of proton and `alpha-` particle.
Given the atomic mass of `"_(1)H^(1)=1.00814 a.m.u.`,
atomic mass of `"_(7)N^(14)=14.00752 amu`, and atomic mass of `"_(2)He^(4)=4.0038 a.m.u.`

Explanation

Applying law of conservation of energy
`(E_(alpha)+m_(alpha)c^(2))+(0+m_(N)c^(2))=(E_(pr)+m_(pr)c^(2))+(E_(x)+m_(x)c^(2))`
where `E` represents the `K.E.`
Energy of the reaction is given by
`Q=KE_(f)=KE_(i)=(E_(pr)+E_(x))-(E_(alpha))=(m_(alpha)+m_(N)-m_(pr)-m_(x))c^(2)`
`rArr -1.26 MeV=(4.00388+14.00752-1.00814-m_(x))xx931.5 MeV`
On solving, we get
`X="_(8)O^(17)`
Applying conservation of linear momentum,
`vecp_(alpha)=vecp_(x)+vecp_(pr)`
`vecp_(alpha)-vecp_(pr)=vecp_(x)`
`p_(alpha)^(2)+p_(pr)^(2)-2p_(alpha)p_(pr)cosphi=p_(x)^(2)`
`:. E_(x)-Q+E_(alpha)-E_(pr)`
`:. E_(x)=(-1.26+7.7-5.5)MeV=0.94`
`:. cos phi= -[(m_(x)E_(x)=m_(alpha)E_(alpha)-m_(pr)E_(pr))/(2sqrt(m_(alpha)m_(pr)E_(alpha)E_(pr)))]=0.4429` (`:'p^(2)=2mE`)
`phi=cos^(-1)(0.4429)=63^(@)42'`