A nucleus of mass `218` amu in Free State decays to emit an `alpha`-particle. Kinetic energy of the `beta-`particle emitted is `6.7 MeV`. The recoil energy (in MeV) of the daughter nucleus is

Options

A. 0.5 MeV
B. 0.25 MeV
C. 0.125 MeV
D. 1.0 MeV

Answer

C

Explanation

Since an `alpha` particle is emitted , the mass number of the daughter nucleus = 218 - 4 = 214 amu.
Using the principle of conservation of linear momentum,
`P_("daughter") = P_alpha`
But `p=sqrt(2Km)` where K is the kinetic energy of the particle .
`therefore sqrt(2K_dm_d) = sqrt(2K_alpham_alpha)`
`therefore K_d m_d = K_alpha m_alpha`
`therefore K_d = (K_alpha m_alpha)/(m_d) = (6.7xx4)/(214) = 1/8 = 0.125 MeV`