A positron is emitted from {}^{23}Na_{11}. Th | vedclass

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Question

(C) During positron emission from a nucleus,a proton is converted into a neutron $(p ightarrow n + e^+ + 
u_e)$.As a result,the atomic number $(Z)$

Vedclass · Accepted Answer

$23 / 10$

Vedclass · Answer

(C) During positron emission from a nucleus,a proton is converted into a neutron $(p ightarrow n + e^+ + 
u_e)$.As a result,the atomic number $(Z)$ decreases by $1$,while the atomic mass $(A)$ remains constant.For the parent nucleus ${}^{23}Na_{11}$,the atomic mass $A = 23$ and atomic number $Z = 11$.After the emission of a positron,the new atomic number $Z' = 11 - 1 = 10$,and the atomic mass $A' = 23$.The ratio of the atomic mass to the atomic number of the resulting nuclide is $\frac{A'}{Z'} = \frac{23}{10}$.Therefore,option $(C)$ is correct.

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

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