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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

What happens to the mass number and atomic number of an element when it emits $\gamma$-radiation?

$\beta$-rays emitted by a radioactive material are

In a radioactive decay process,the atomic number and mass number of the nucleus do not change. Which of the following is emitted during this process?

The radioactive nucleus $_7N^{13}$ decays to $_6C^{13}$ through the emission of

${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are

Vedclass Test Series

Exam Paper Generator

Online Exam Module