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# Calculate the energy released in the following reaction, given the masses to be\\ce{_6^11C -> _5^11B + e^+ + neutrino}\ - Physics

Numerical

Calculate the energy released in the following reaction, given the masses to be

$\ce{_88^223Ra}$ : 223.0185 u, $\ce{_82^209Pb}$ : 208.9811 u, $\ce{_6^14C}$ : 14.00324 u, $\ce{_92^236U}$ : 236.0456 u, $\ce{_56^140Ba}$ : 139.9106 u, $\ce{_36^94Kr}$ : 93.9341 u, $\ce{_6^11C}$ : 11.01143 u, $\ce{_5^11B}$ : 11.0093 u. Ignore neutrino energy.

$\ce{_6^11C -> _5^11B + e^+ + neutrino}$

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#### Solution

$\ce{_6^11C -> _5^11B + e^+ + neutrino}$

The energy released in this reaction = (ΔM)c2

= [11.01143 - (11.0093 + 0.00055)](931.5} MeV

=1.47177 MeV

Concept: Nuclear Energy
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 15.3 | Page 343
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