The isotope `_(5)^(12) B` having a mass `12.014 u`undergoes beta - decay to `_(6)^(12) C _(6)^(12) C `has an excited state of the nucleus `( _(6)^(12) C ^(**) at 4.041 MeV` above its ground state if `_(5)^(12)E` decay to `_(6)^(12) C ^(**) ` , the maximum kinetic energy of the `beta` - particle in unit of MeV is `(1 u = 931.5MeV//c^(2)` where c is the speed of light in vaccuum) .

Options

A. `9`
B. `6`
C. `3`
D. `1`

Answer

A

Explanation

`._(5)B^(12)to._(6)C^(12)+._(-1)e^(0)+barv`
Taking mass of `c^(12)` as 12.000u, mass defect
`Deltam=12.014-12.000=0.014u`
`Q=0.014xx931.5MeV =13.041 MeV`
Out of this, 4.041MeV energy is taken by `._(6)C^(12**)`
`:.` Max. K.E. of `beta` particle=13.041-4.041
`=9MeV`