Question

Number of neutrons in a parent nucleus X, which gives ${}_7{N}^{14}$ two successive $\beta$ emission, would be :

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The correct option is D 9

Beta emission means conversion of neutron into protons so
${}_Z{X}^A\to {}_7{N}^{14}+2{}_{-1}{\beta }^0$.
Hence, A = 14, Z = 5
Neutrons = 14 - 5 = 9