IIT-JEE 2007 · Paper 2 · Chemistry Q29

ChemistrySingle correct+3 / -1Question 29 of 66

A positron is emitted from $\ce{^{23}_{11}Na}$ . The ratio of the atomic mass and atomic number of the resulting nuclide is

22/10

22/11

23/10

23/12

Answer and worked solution

Official answer: C

Positron emission turns a proton into a neutron: the mass number stays 23 and Z falls by 1:
$\ce{^{23}_{11}Na -> ^{23}_{10}Ne + e+ + \nu}.$

Ratio $23/10$ .

(C)

Topics:Radioactive decay kinetics

IIT-JEE 2007 · Paper 2, Chemistry question 29