Question 10.2: Write a balanced nuclear equation for the β emission of phos...
Write a balanced nuclear equation for the β emission of phosphorus-32, a radioisotope used to treat leukemia and other blood disorders.
Analysis
Balance the atomic numbers and mass numbers on both sides of a nuclear equation. With β emission, treat the β particle as an electron with zero mass in balancing mass numbers and a − 1 charge when balancing the atomic numbers.
[1] Write an incomplete equation with the original nucleus on the left and the particle emitted on the right.
• Use the identity of the element to determine the atomic number; phosphorus has an atomic number of 15.
{ }_{15}^{32} \mathrm{P} \longrightarrow{ }_{-1}^0 \mathrm{e}+\text { ? }
[2] Calculate the mass number and the atomic number of the newly formed nucleus on the right.
• Mass number: Since a β particle has no mass, the masses of the new particle and the original particle are the same, 32.
• Atomic number: Since β emission converts a neutron into a proton, the new nucleus has one more proton than the original nucleus; 15 = − 1 + ?. Thus the new nucleus has an atomic number of 16.
[3] Use the atomic number to identify the new nucleus and complete the equation.
• From the periodic table, the element with an atomic number of 16 is sulfur, S.
• Write the mass number and the atomic number with the element symbol to complete the equation.
{ }_{15}^{32} \mathrm{P} \longrightarrow{ }_{-1}^0 \mathrm{e}+{ }_{16}^{32} \mathrm{~S}
