A radioactive nuclide used for radiation therapy for cancer has 33 neutrons and a mass number of 60. Which is the correct symbol for this nuclide?

A. [tex]\(_{60}^{27} Co\)[/tex]
B. [tex]\(_{60}^{33} Co\)[/tex]
C. [tex]\(_{33}^{60} Co\)[/tex]
D. [tex]\(_{27}^{60} Co\)[/tex]



Answer :

To determine the correct symbol for the nuclide used in radiation therapy, we need to understand how to write the chemical symbol for an element based on its number of protons (atomic number) and mass number.

Given:
- The number of neutrons in the nuclide is 33.
- The mass number of the nuclide is 60.

Recall that the mass number (A) is the sum of the number of protons (Z) and neutrons (N) in the nucleus:
[tex]\[ A = Z + N \][/tex]

We can rearrange this formula to find the number of protons:
[tex]\[ Z = A - N \][/tex]

Substituting the given values:
[tex]\[ Z = 60 - 33 \][/tex]
[tex]\[ Z = 27 \][/tex]

The element with atomic number (Z) 27 is Cobalt (Co). Therefore, the symbol for the nuclide is given in the format:
[tex]\[ {}^{\text{mass number}}_{\text{atomic number}} \text{element symbol} \][/tex]

Thus, the correct symbol for this nuclide is:
[tex]\[ {}^{60}_{27} \text{Co} \][/tex]

So, the correct option is:
[tex]\[ {}^{60}_{27} \text{Co} \][/tex]