Find the energy liberated in the reaction 223 Ra → 209 Pb + 14 C. The atomic masses needed are as follows. 223 Ra 209 Pb 14 C 223.018 u 208.981 u 14.003 u

Question

Philoid · Accepted Answer

M_Pb = mass of Pb²⁰⁹=208.981 u

M_C = mass of C¹⁴ =14.003 u

M_Ra = mass of Ra²²³=223.018 u

The energy released is 31.671 MeV

Find the energy liberated in the reaction223Ra → 209Pb + 14C.The atomic masses needed are as follows.223Ra 209Pb 14C223.018 u 208.981 u 14.003 u

Find the energy liberated in the reaction
²²³Ra → ²⁰⁹Pb + ¹⁴C.

The atomic masses needed are as follows.

²²³Ra ²⁰⁹Pb ¹⁴C

223.018 u 208.981 u 14.003 u