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The radionuclide .6^11C decays according...

The radionuclide `._6^11C` decays according to `._6^11C to ._5^11B + e^(+) + v`. `T_(1//2)` = 20.3 min. The maximum energy of emitted positron is 0.960 MeV. Calculate Q and compare it with the maximum energy of the positron emitted. Given : `m(._6^11C)` =11.01143 u, `m(._5^11B)` =11.009305 u, `m_e` = 0.000548 u

Text Solution

Verified by Experts

`Q=m_N(._6^11C)-[m_N(._5^11B)+ m_e]`
[`m_N` stands for the nuclear mass of the element or particle =atomic mass - mass of extranuclear electrons ]
`therefore Q=[m(._6^11C) -6m_e] -[m(._5^11B) -5m_e + m_e]`
`=m(._6^11C)-m (._5^11B)-2m_e`
=(11.011434-11.009305-2 x 0.000548) u
=0.001033 x 931.2 MeV =0.962 MeV
This energy is almost equal to the maximum energy released in the decay process.
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