Chapter 19, Problem 8STP

Interpretation Introduction

Interpretation: The balanced nuclear equation for the decay of $\text{P}$ to produce a positron has to be written.

Concept Introduction: The emission of positron $\left({\text{e}}^{\text{+}}\right)$ on decay of a nucleus is called as positron emission. In positron emission, one proton of original nucleus decays to a form positron and a neutron.

The general equation is,

  $X\to e+Y$

Answer to Problem 8STP

The balanced nuclear equation for the decay of $\text{P}$ to produce a positron is,

  $\text{P}\to e+\text{S}\text{i}$

Explanation of Solution

The decay of a nucleus by emitting a positron $\left({\text{e}}^{\text{+}}\right)$ is called as positron emission. In positron emission, one proton of original nucleus decays to a form positron and a neutron. Due to which, the new nucleus formed has Z-1 protons and A-Z+1 neutrons.

The general equation is,

  $X\to e+Y$

Where,

• $A$is the mass number.
• $Z$ is the atomic number.
• $X$ is the original isotope.
• $Y$ is the new isotope.

Positron emission converts a proton into a neutron, the new nucleus formed with have one fewer proton than the original nucleus but the mass of the original and new particle will be the same.

In phosphorus $\left(\text{P}\right),$ the mass number is $30$ and the atomic number is $15.$

Mass number: Since a positron particle has no mass, the masses of the new and original particle could be the same $\left(\text{30}\right)\text{.}$

Atomic number: Since positron particle converts a proton into a neutron, the new nucleus would have Z-1 protons, $\text{15-1=14}\text{.}$ Therefore, the new nucleus has atomic number of $\text{14}\text{.}$

From the periodic table, the element that has atomic number of $14$ is identified as silicon.

Phosphorus undergoes positron emission to form silicon $\left(\text{S}\text{i}\right),$ whose atomic number will be $14$ , the mass number will be $30$ and positron would be emitted.

The balanced nuclear equation of $\text{P}$ to produce a positron is,

  $\text{P}\to e+\text{S}\text{i}$

Conclusion

In position emission, the atomic number of the original nucleus is Z-1and the mass number is A. The balanced nuclear equation of $\text{P}$ to produce a positron is,

  $\text{P}\to e+\text{S}\text{i}$