The boron14 nucleus mass 1402266 u beta decays spontaneously

The boron-14 nucleus (mass: 14.02266 u) “beta decays”, spontaneously becoming an electron (mass: 0.00055 u) and a carbon-14 nucleus (mass: 13.99995 u). What will be the speeds and kinetic energies of the carbon-14 nucleus and the electron? (note: we are ignoring the production of a neutrino. In this problem the carbon nucleus recoils “slowly” so you can assume u =1 for the carbon. Lastly recall that momentum and energy will both be conserved in this decay.)

Solution

before decay :

mass = 14.02266 u

after dacay mass = 0.00055 + 13.99995 = 14.00050 u

change in mass = 14.02266 - 14.00050 = 0.02216 u

energy of system after decay = 931.5 MeV / u x 0.02216 u = 20.64 MeV

    = 20.64 x 10^6 x 1.6 x 10^-19 J = 3.3024 x 10^-12 J
Using momentum conservation,

initial momentum = final momentum

0 = MV - mv

13.99995V = 0.00055v

v = 25454.45V

final KE of system,

M V^2 /2 + mv^2 /2 = 3.3024 x 10^-12 J

(13.99995 x 1.66 x 10^-27 x V^2)/2 + (0.00055 x 1.66 x 10^-27 x (25454.45V)^2 ) /2 = 3.3024 x 10^-12

5.9158 x 10^-17 v^2 = 6.6048 x 10^-12

v = 1.06 x 10^5 m/s ...........Speed of electron,

speed of carbon-14 nucleus , V = (1.06 x 10^5) / 25454.45 = 4.15 m/s .....Ans

KE of electron = (0.00055 x 1.66 x 10^-27 x (25454.45x 4.15)^2 ) /2

       = 5.10 x 10^-21 J

KE of carbon-14 nucleus = 13.999955 x 1.66 x 10^-27 x 4.15^2 /2

   = 2 x 10^-25 J