Question:
\(^{14}\)N + \(^{1}_{0}\)n \(\rightarrow\) \(^{14}_{6}\)C + \(^{1}_{1}\)H is written as
\(^{14}\)N + \(^{1}_{0}\)n \(\rightarrow\) \(^{14}_{6}\)C + \(^{1}_{1}\)H is written as
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In nuclear notation: (projectile, ejected particle).
Updated On: Apr 23, 2026
- \(^{14}\)N(n, e)\(^{1}_{1}\)H
- \(^{14}\)N(p, n)\(^{14}_{6}\)C
- \(^{14}\)N(n, p)\(^{14}_{6}\)C
- \(^{16}\)C(p, n)\(^{14}_{7}\)N
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Nuclear reaction notation: Target(projectile, ejected)product.
Step 2: Detailed 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: Final Answer:
Thus, correct notation is \(^{14}\)N(n, p)\(^{14}_{6}\)C.
Nuclear reaction notation: Target(projectile, ejected)product.
Step 2: Detailed 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: Final Answer:
Thus, correct notation is \(^{14}\)N(n, p)\(^{14}_{6}\)C.
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