the radioactive atom 61 27 co emits a beta particle write an equation showing the decay

 when a radioactive atom emits a beta particle , one of its neutron decays into a proton and an electron , an electron neutrino is also emitted.

 The atomic number increases by one while the mass number remain unchanged. The new atom formed is more positive than the original atom, in other word the daughter atom has more protons than the parent atom.

 for the beta decay of cobalt-60, we can have the following equation

⁶⁰₂₇Co→ X +⁰_-1e

To find the daughter nucleus  we do this.

 The mass number does not

 Therefore the mass remain the same

 The atomic number increases by one

 Therefore we say the atomic number of cobalt, 27 + 1=28

 We go back to the periodic table to find what atomic has the atomic number of 28.

 The atom that fit the above description is nickel-60.

 Therefore the above equation  will be

⁶⁰₂₇Co→ ⁶⁰₂₈Ni +⁰_-1e

Let us check if the above equation is balanced

 To balance a nuclear equation, we observe two criteria

 The mass number on the product side must be equal to the mass number on the reactant side.

 The atomic number on the product side must be equal to the atomic number on the reactant side.

Mass numbers

 On the reactant side we have 60  and on the  product side we have 60 +0 =60.

Atomic number

On the reactant side we have 27 and on the product side we have 27 + (-1)=27

 Therefore the equation is balanced and we are not wrong.