During alpha-decay , a nucleus decays by emitting an `alpha`-particle ( a helium nucleus `._2He^4`) according to the equation
`._Z^AX to ._(Z-2)^(A-4)Y+._2^4He+Q`
In this process, the energy released Q is shared by the emitted `alpha`-particle and daughter nucleus in the form of kinetic energy .
The energy Q is divided in a definite ratio among the `alpha`-particle and the daughter nucleus .
A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo `beta`-decay .This process also involves a release of definite energy . Initially, the `beta`-decay was represented as `._Z^AX to ._(Z+1)^AY + e^(-)"(electron)"+Q`
According to this reaction, the energy released during each decay must be divided in definite ratio by the emitted e' (`beta`-particle) and the daughter nucleus. While , in alpha decay, it has been found that every emitted `alpha`-particle has the same sharply defined kinetic energy. It is not so in case of `beta`-decay . The energy of emitted electrons or positrons is found to vary between zero to a certain maximum value. Wolfgang Pauli first suggested the existence of neutrinoes in 1930. He suggested that during `beta`-decay, a third particle is also emitted. It shares energy with the emitted `beta` particles and thus accounts for the energy distribution.
During `beta^+` decay (positron emission) a proton in the nucleus is converted into a neutron, positron and neutrino. The reaction is correctly represented as

Options

A. `._Z^AX to ._(Z-1)^(A-1)X + e^(+) + v`+ Energy
B. `._Z^AX to ._(Z-1)^(" "A) X + e^(+) + v`+ Energy
C. `._Z^AX to ._(Z-1)^A X + e^(+) + v`+ Energy
D. `._Z^AX to ._Z^(A-1)X + e^(+) +v +` Energy

Answer

C