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Sahay Sir > Question Answers > Cengage JEE Mains Physics by B.M Sharma > Chapter 30 - Nuclear Physics > Stationery nucleus .238U decays by a emission generating a total kinetic energy T: .23892U→.23490Th+.42α What is the kinetic energy of the α-particle?

Stationery nucleus .238U decays by a emission generating a total kinetic energy T: .23892U→.23490Th+.42α What is the kinetic energy of the α-particle?

Stationery nucleus .238U decays by a emission generating a total kinetic energy T: .23892U→.23490Th+.42α What is the kinetic energy of the α-particle?