Question

The decay modes of \( ^{14}C \) and \( ^{14}O \) are

• A. \( \beta \) decay
• B. positron emission
• C. \( \beta \) decay and positron emission, respectively
• D. positron emission and \( \beta \) decay, respectively

Hint:

The decay modes of isotopes are determined by the ratio of neutrons to protons. \( \beta \)-decay occurs when there are too many neutrons, and positron emission occurs when there are too many protons.

Answer 1

The Correct Option is C

Step 1: Understanding the decay modes.
Carbon-14 (\( ^{14}C \)) undergoes \( \beta \)-decay, where a neutron is converted into a proton, emitting an electron (beta particle). On the other hand, oxygen-14 (\( ^{14}O \)) undergoes positron emission, where a proton is converted into a neutron, emitting a positron.

Step 2: Analyzing the options.
(A) \( \beta \) decay: Incorrect — \( ^{14}C \) undergoes \( \beta \)-decay, but \( ^{14}O \) does not.
(B) Positron emission: Incorrect — \( ^{14}C \) undergoes \( \beta \)-decay, not positron emission.
(C) \( \beta \) decay and positron emission, respectively: Correct — This correctly describes the decay modes of \( ^{14}C \) and \( ^{14}O \).
(D) Positron emission and \( \beta \) decay, respectively: Incorrect — This order is reversed for the two isotopes.

Step 3: Conclusion.
The correct answer is (C) as \( ^{14}C \) undergoes \( \beta \)-decay, and \( ^{14}O \) undergoes positron emission.