13.12 Find the Q-value and the kinetic energy of the emitted -particle in the -decay of (a) and (b) . Given m ( = 226.02540 u, m ( = 222.01750 u, m ( = 220.01137 u, m ( = 216.00189 u.
13.12 Find the Q-value and the kinetic energy of the emitted -particle in the -decay of (a) and (b) . Given m ( = 226.02540 u, m ( = 222.01750 u, m ( = 220.01137 u, m ( = 216.00189 u.
13.12 particle decay of emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.
+
Q value of emitted particle = (Sum of initial mass – Sum of final mass) , where
c = Speed of light.
It is given that
m ( = 226.02540 u
m ( = 222.01750
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Physics Ncert Solutions Class 12th 2023
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