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InterviewSolution
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A radioactive element decays by `beta-` emission. If mass of parent and daughter nucliede are `m_(1)` and `m_(2)` respectively, caluclate energy liberated during the emission. |
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Answer» The masses of parent and daughter elements nuclide are `m_(1)` and `m_(2)` respectively. `._(2)^(X)A rarr _(Z + 1)^(X)B + ._(-1)^(0)e` Mass of parent element `= m_(1) + Zm_(e)` (where `Z` is at. No. of parent element.) Mass of daughter element `= m_(2) + (Z + 1)m_(e)` (As a. no fo element increases by one due to `beta-` emission) `:.` Mass decay = Mass of parent atom - One `beta` loss- [Mass of daugher atom along with one electron] `= m_(1) Zm_(e) - m_(e) - [m_(2) + (Z + 1) m_(e)]` `= [m_(1) - m_(2) - 2m_(e)]` Energy liberated `= [m_(1) - m_(2) - 2m_(e)] c^(2)` |
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