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In the following nuclear reaction, Identify unknown labelled X.\(^{22}_{11}Na+X\rightarrow ^{22}_{10}Ne+v_e\) |
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Answer» \(^{22}_{11}Na\) + \((^o_{-1}e^-)\) → \(^{22}_{10}Ne\) + \(\bar{v}_e\)(electron antineutrino) \(^o_{-1}e\) (or \(^o_{-1}\beta)\) \(\Rightarrow\) beta particle (electron) The answer is: Electron. |
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