Question

1. show that the mass number and the number of protons are conserved in the following nuclear equation:

_{92}^{234}u \to _{90}^{230}th + _{2}^{4}he.

Explanation:

Step1: Calculate mass number on both sides

Mass number of \(^{234}_{92}U\) is \(234\).
Mass number of \(^{230}_{90}Th\) is \(230\) and of \(^{4}_{2}He\) is \(4\).
Sum of mass numbers on right - hand side: \(230 + 4=234\)

Step2: Calculate number of protons on both sides

Number of protons in \(^{234}_{92}U\) is \(92\).
Number of protons in \(^{230}_{90}Th\) is \(90\) and in \(^{4}_{2}He\) is \(2\).
Sum of protons on right - hand side: \(90+2 = 92\)

Answer:

Since the mass number (\(234\)) and the number of protons (\(92\)) are the same on both sides of the nuclear equation \(^{234}_{92}U
ightarrow^{230}_{90}Th + ^{4}_{2}He\), the mass number and the number of protons are conserved.