13.12 Find the Q-value and the kinetic energy of the emitted α -particle in the α -decay of (a) Ra88226 and (b) Rn86220 . Given m ( Ra88226) = 226.02540 u, m ( Rn86222) = 222.01750 u, m ( Rn86222) = 220.01137 u, m ( Po)84216 = 216.00189 u.

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7 months ago

13.12α- particle decay of Ra88226 emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

Ra88226  Rn86222 + He24

Q value of emitted α- particle = (Sum of initial mass – Sum of final mass) *c2 , where

c = Speed of light.

It is given that

m ( Ra88226) = 226.02540 u

m ( Rn86222) = 222.01750

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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