Nuclei `X` decay into nuclei `Y` by emitting `alpha`-aprticles. Energies of `alpha`-particle are found to be only `1 MeV` & `1.4 MeV`. Disregarding the recoil of nuclei `Y`. The energy of `gamma` photon emitted will be

A nucleus of mass 220 amu in the free state decays to emit an alpha -particle . Kinetic energy of the alpha -particle emitted is 5.4 MeV. The recoil energy of the daughter nucleus is

A nucleus of mass 220 amu in the free state decays to emit an alpha -particle . Kinetic energy of the alpha -particle emitted is 5.4 MeV. The recoil energy of the daughter nucleus is

A nucleus ""^(220)X at rest decays emitting an alpha -particle . If energy of daughter nucleus is 0.2 MeV , Q value of the reaction is

A nucleus ""^(220)X at rest decays emitting an alpha -particle . If energy of daughter nucleus is 0.2 MeV , Q value of the reaction is

A high-energy gamma-photon may turn into an electron-positron pair in the field of heavy nuclei. What is the minimum energy of the gamma-photon?

The alpha -decay of nucleus X^(230) is accompanied by emission of two groups of alpha- particles-one group having alpha- particles of kinetic energy 6.7MeV and other having alpha -particles of kinetic energy 5.2 MeV. Following the emission of these particles the daugheter muclei are found in ground and exiceted states. find the energy of gamma photon. that is subsequenctly emitted by de-exictation of the excited daughter nuclei. Assume that nuclei of X^(230) are at rest.