Question

Calculate the energy released in the following reactions, given the masses to be \(^{223}_{88}Ra\) : 223.0185 u,\(^{209}_{82}Pb\) : 208.9811, \(^{14}_{6}C\): 14.00324, \(^{236}_{92}U\): 236.0456, \(^{140}_{56}Ba\): 139.9106, \(^{94}_{36}Kr\): 93.9341, \(^{11}_{6}C\): 11.01143, \(^{11}_{5}B\): 11.0093. Ignore neutrino energy.

(a) \(^{223}_{88}Ra\) \(\longrightarrow\) \(^{209}_{82}Pb\) + \(^{14}_{6}C\)

(b) \(^{236}_{92}U\) \(\longrightarrow\) \(^{140}_{56}Ba\) + \(^{94}_{36}Kr\) + 2n

(c) \(^{11}_{6}C\) \(\longrightarrow\) \(^{11}_{5}B\) + e⁺ + neutrino

Answer 1

Data : \(^{223}_{88}Ra\) : 223.0185u, \(^{209}_{82}Pb\) : 208.9811u,

(a) \(^{223}_{88}Ra\) \(\longrightarrow\) \(^{209}_{82}Pb\) + \(^{14}_{6}C\)

The energy released in this reaction = (∆M) c² = [223.0185 – (208.9811 + 14.00324)j(931 .5) MeV

= 31.820004 MeV

(b) \(^{236}_{92}U\) \(\longrightarrow\) \(^{140}_{56}Ba\) + \(^{94}_{36}Kr\) + 2n

The energy released in this reaction = (∆M) c² = [236.0456 – (139.9106 + 93.9341 + (2)(1 .00866)1(93 1 .5)MeV 

= 171.00477 MeV

(c) \(^{11}_{6}C\) \(\longrightarrow\) \(^{11}_{5}B\) + e⁺ + neutrino

The energy released in this reaction = (∆M) c² = [11.01143 – (11.0093 + O.00055)](931.5) MeV 

= 1.47177 MeV