\(^{114}_{55}Cs \rightarrow ^{A}_{Z}E + ^{4}_{2}\alpha\) Find the value of A and Z in the equation...
\(^{114}_{55}Cs \rightarrow ^{A}_{Z}E + ^{4}_{2}\alpha\)
Find the value of A and Z in the equation above
- A) 119, 53
- B) 110, 57
- C) 110, 53
- D) 110, 58
Correct Answer: C) 110, 53
Explanation
This question involves a nuclear decay reaction, specifically alpha decay. In an alpha decay, an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. The symbol for an alpha particle is \(^{4}_{2}\alpha\).
Given the decay reaction: \[^{114}_{55}Cs \rightarrow ^{A}_{Z}E + ^{4}_{2}\alpha\]
We are asked to find the values of A and Z for the element E that is formed as a result of the decay.
To find the values of A and Z, we need to balance the nuclear equation. This means that the number of protons (atomic number) and the number of nucleons (mass number) should be conserved during the reaction.
For the atomic number (Z):
\[55 = Z + 2\]
\[Z = 55 - 2 = 53\]
For the mass number (A):
\[114 = A + 4\]
\[A = 114 - 4 = 110\]
So, the decay reaction is: \[^{114}_{55}Cs \rightarrow ^{110}_{53}E + ^{4}_{2}\alpha\]
The values of A and Z are 110 and 53, respectively. Thus, the correct option isOption C: 110, 53.