Question

\(^{14}\)N + \(^{1}_{0}\)n \(\rightarrow\) \(^{14}_{6}\)C + \(^{1}_{1}\)H is written as

• A. \(^{14}\)N(n, e)\(^{1}_{1}\)H
• B. \(^{14}\)N(p, n)\(^{14}_{6}\)C
• C. \(^{14}\)N(n, p)\(^{14}_{6}\)C
• D. \(^{16}\)C(p, n)\(^{14}_{7}\)N

Hint:

In nuclear notation: (projectile, ejected particle).

Answer 1

The Correct Option is C

Step 1: Concept:
Nuclear reaction notation: Target(projectile, ejected)product.
Step 2: Explanation:
Reaction: \(^{14}\)N + \(^{1}_{0}\)n \(\rightarrow\) \(^{14}_{6}\)C + \(^{1}_{1}\)H.
Target = \(^{14}\)N, projectile = n, ejected particle = p (\(^{1}_{1}\)H), product = \(^{14}_{6}\)C.
Notation: \(^{14}\)N(n, p)\(^{14}_{6}\)C.
Step 3: Conclusion:
Thus, correct notation is \(^{14}\)N(n, p)\(^{14}_{6}\)C.