Question

When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?

Answer 1

That will make a gold-202 nucleus.

Explanation

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the $\displaystyle ^{202}_{\phantom{2}80}\text{Hg}$ with a neutron $^{1}_{0}n$. The neutron will add 1 to the mass number 202 of $^{202}_{\phantom{2}80}\text{Hg}$. However, the atomic number will stay the same.

• New mass number: 202 + 1 = 203.
• Atomic number is still 80.

$^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}$.

Double check the equation:

• Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
• Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The $^{203}_{\phantom{2}80}\text{Hg}$ nucleus loses a proton $^{1}_{1}p$. Both the mass number 203 and the atomic number will decrease by 1.

• New mass number: 203 - 1 = 202.
• New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

$^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p$.

Double check the equation:

• Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
• Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.