\(^{226}_{88}Ra \rightarrow ^{x}_{86}Rn + \alpha\) What is the value of x in the nuclear reaction...
\(^{226}_{88}Ra \rightarrow ^{x}_{86}Rn + \alpha\) What is the value of x in the nuclear reaction above?
- A) 220
- B) 222
- C) 226
- D) 227
Correct Answer: B) 222
Explanation
The given nuclear reaction involves the decay of a radium (Ra) isotope into a radon (Rn) isotope and an alpha particle (?). In this type of decay, an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons.
In the reaction, the radium isotope is represented as \(^{226}_{88}Ra\), where 226 is the mass number (total number of protons and neutrons) and 88 is the atomic number (number of protons). The radon isotope is represented as \(^{x}_{86}Rn\), where x is the mass number we need to determine, and 86 is the atomic number.
An alpha particle (?) consists of 2 protons and 2 neutrons, so its mass number is 4, and its atomic number is 2. When the radium decays, the mass number and atomic number must be conserved, meaning the sum of the mass numbers and atomic numbers on both sides of the reaction must be equal.
For mass numbers:
\(226 = x + 4\)
To find the value of x, subtract 4 from 226:
\(x = 226 - 4\)
\(x = 222\)
So, the value of x in the nuclear reaction is222, makingOption B the correct answer.