A nucleus at rest undergoes a decay emitting an `alpha` -particle of de-broglie wavelength `lambda = 5.76 xx 10^(-15) m`. The mass of daughter nucleus is 223.40 amu and that of `alpha`-particle is 4.002 amu - The linear momentum of `alpha` -particle and that of daughter nucleous is -

A nucleus at rest undergoes a decay emitting an alpha -particle of de-broglie wavelength lambda = 5.76 xx 10^(-15) m . The mass of daughter nucleus is 223.40 amu and that of alpha-particle is 4.002 a.m.u.detrmine the total kinetic energy in the final state.Hence,obtain teh mass of the parent nucleus in a.m.u. Given that 1 a.m.u.=931.470 MeV .

A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength lambda = 5.76 xx 10^(-15)m if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is 4.002amu , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 MeV//e^(2) )

A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength lambda = 5.76 xx 10^(-15)m if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is 4.002amu , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 MeV//e^(2) )

A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength lambda = 5.76 xx 10^(-15)m if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is 4.002amu , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 MeV//e^(2) )

A nuclear af rest undergoes a decay emitting and alpha- particle of de-Broglie wavelength lamda=5.76xx10^(-15) meter. he mass of the daughter nucleus is 223.610 amu and that of the alpha - particle is 4.002 a.m.u. (1"amu"=931.470(MeV)/(C^(2))) Momentum of the alpha - particle is nearly