A nucleus with mass number 220 initially at rest emits an `alpha` particle. If the Q value of the reaction is 5.5 MeV , the kinetic energy of the `alpha` particle is

Options

A. `4.4 MeV`
B. `5.4 MeV`
C. `5.6MeV`
D. `6.5MeV`

Answer

b

Explanation

By consevation of linear momentum,
`MV=mv rArr 126 V=4v rArr V=(v)/(54)`
By energy conservation,

`(1)/(2) mv^(2)+(1)/(2)MV^(2)=5.5xx1.6xx10^(-13) J`
`rArr 4 xx 1.67 xx 10^(-27) v^(2) + 216 xx 1.67 xx 10^(-27) xx (v^(2))/(54 xx 54)`
`= 2 xx 5.5 xx 1.6 xx 10^(-13)`
`rArr 1.67 xx 10^(-27) v^(2) [4+(216)/(54 xx 54)]`
`=2 xx 5.5 xx 1.6 xx 10^(-13)`
or `v^(2) = (2 xx 5.5 xx 1.6 xx 54 xx 54 xx 10^(-13))/((4 xx 54 xx 54 + 216)1.67 xx 10^(-27))` ,brgt `= 2.586 xx 10^(14)`
`:.` KE of `alpha`-particle `= (1)/(2) mv^(2)`
`= (1)/(2) xx 4 xx 1.67 xx 10^(-27) xx 2.586 xx 10^(14)`
`= 8.637 xx 10^(-13)J = (8.637 xx 10^(-13))/(1.6 xx 10^(-13)) Me V`
`= 5.4 Me V`
Alternative solution :
By conservation of momentum,
`P_(1) = P`
`rArr sqrt(2k_(2) m_(1)) = sqrt(2k_(2) m_(2))`
`rArr sqrt(2k_(1) (216)) = sqrt(2k_(2) (4))`
`rArr k_(2) = 54 k_(1)` ...(i)
Also, `k_(1) + k_(2) = 5.5 MeV` ...(ii)
Solve Eqs. (i) and (ii).