Question:

A positron is emited from  23Na_11 . The ratio of the atomic mass and atomic number of the resulting nuclide is:

(a) 23/10

(b) 22/10

(c) 23/12

(d) 22/11

Solution

Positron emission or beta plus decay (beta+ decay) is a type of radioactive decay in which a proton inside a nucleus is converted into a neutron while releasing a positron and an electron neutrino (V_e)

 1p_1 rarr 1n_0 + 0e_1 +V_e

0e_1 +V_e are released and proton inside the atom turns into a neutron.

 23Na_11 rarr 23Na_10 + e^+ +V_e

So atomic number decreases by one but atomic mass remains constant.
Atomic mass/atomic number = 23/10

Therefore, option (a) is correct.