\(^{226}_{88}Ra \rightarrow ^x_{86}Rn + \alpha - particle\). What is the value of x in the...
\(^{226}_{88}Ra \rightarrow ^x_{86}Rn + \alpha - particle\). What is the value of x in the nuclear reaction?
- A) 226
- B) 220
- C) 222
- D) 174
Correct Answer: C) 222
Explanation
In the given nuclear reaction: \[^{226}_{88}Ra \rightarrow ^x_{86}Rn + \alpha - particle\]
An alpha particle is a helium nucleus, which consists of 2 protons and 2 neutrons. In terms of atomic and mass numbers, an alpha particle can be represented as \[^{4}_{2}He\].
To find the value of x in the nuclear reaction, we need to balance the atomic and mass numbers on both sides of the reaction.
For the atomic numbers (bottom numbers), we have:
\(88 = 86 + 2\)
For the mass numbers (top numbers), we have:
\(226 = x + 4\)
To solve for x, we can rearrange the equation:
\(x = 226 - 4\)
Thus, the value of x is:
\(x = 222\)
So, the correct answer isOption C: 222. This nuclear reaction represents the radioactive decay of radium-226 into radon through the emission of an alpha particle.