Question

Consider the following nuclear fission reaction
\({}_{0}^{1}n+{}_{92}^{235}U\rightarrow {}_{Z}^{A}Y+{}_{41}^{99}Nb+neutrons\).
If \(4\) neutrons are released during the fission process, then \(A\) and \(Z\) are

• A. \(A=133\) and \(Z=51\)
• B. \(A=51\) and \(Z=133\)
• C. \(A=339\) and \(Z=133\)
• D. \(A=133\) and \(Z=339\)

Hint:

Mass number and atomic number must both be conserved on the two sides of the nuclear reaction.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a nuclear reaction, the total mass number A and total atomic number Z are conserved. Neutrons have Z = 0.

Step 2: Balance mass number
\(1+235=A+99+4\times 1\)
\(236=A+103\Rightarrow A=133\)

Step 3: Balance atomic number
\(0+92=Z+41+0\)
\(Z=51\)

Final Answer:
The fragment has A equal to 133 and Z equal to 51. \[ \boxed{\text{(A) }A=133\text{ and }Z=51} \]