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13.Nuclei
hard
A positron is emitted from ${ }^{23} \mathrm{Na}_{11}$. The ratio of the atomic mass and atomic number of the resulting nuclide is
A
$22 / 10$
B
$22 / 11$
C
$23 / 10$
D
$23 / 12$
(IIT-2007)
Solution
On position emission from nucleus, proton converts into neutron therefore atomic number decreases by one but atomic mass remains constant.
$\frac{\text { Atomic mass }}{\text { atomic number }}=\frac{23}{10}$
Hence $(\mathrm{C})$ is correct.
Standard 12
Physics
