A nucleus of mass number 220 decays by `alpha` decay. The energy released in the reation is 5 MeV. The kinetic energy of an `alpha-` particle is

Options

A. `(1)/(54)MeV`
B. `(27)/(11)MeV`
C. `(54)/(11)MeV`
D. `(55)/(54)MeV`

Answer

C

Explanation

Let the reaction be represented as
`{:("220 216 4"),(" "X " "rarr " "Y+He),("Z Z-2 2"):}`
`because` energy released in the reaction is 5 MeV
`rArr (1)/(2)m_(gamma)upsilon_(Y)^(2)+(1)/(2)m_(alpha)upsilon_(alpha)^(2)=5MeV`
`rArr (1)/(2)(216)upsilon_(gamma)^(2)+(1)/(2)4(upsilon_(alpha))^(2)=5MeV" ...(i)"`
Also using conservation of linear momentum.
`m_(gamma)upsilon_(gamma)=-m_(alpha)upsilon_(alpha) rArr upsilon_(gamma)=(-m_(alpha))/(m_(gamma))upsilon_(alpha)`
`upsilon_(gamma)=(-4)/(216)upsilon_(alpha)=(-1)/(54)upsilon_(alpha)`
Putting in eqn. (i)
`(1)/(2)(216)((upsilon_(alpha))/(54))^(2)+(1)/(2)(upsilon_(alpha))^(2)="5 MeV"`
`(1)/(2)(216)((upsilon_(alpha))/(54))^(2)+K.E._(alpha)="5 MeV"`
`rArr (1)/(2)xx4(upsilon_(alpha)^(2))/(54)+"K.E."_(alpha)="5 MeV"`
`"K.E."_(alpha)=(5xx54)/(55)=(54)/(11)"MeV"`