Question Detail

CHM 102 Radioactivity Objective Question
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Question

Complete the following nuclear reaction from the document: 147N + 42He → __________ + 11p
Options
A <sup>17</sup><sub>8</sub>O
Correct Answer
B <sup>13</sup><sub>6</sub>C
C <sup>14</sup><sub>6</sub>C
D <sup>15</sup><sub>7</sub>N
Correct Answer

Option A is the correct answer.

Detailed Explanation

Page 7 lists this reaction as '147N + 42He → 178O + 11p'. Sum of mass numbers: 14 + 4 = 18. Sum of atomic numbers: 7 + 2 = 9. For conservation, the missing nuclide must have a mass number of 18 - 1 = 17 and an atomic number of 9 - 1 = 8, corresponding to Oxygen-17 (178O).

Hint

The sum of mass numbers and the sum of atomic numbers must be conserved on both sides of the nuclear equation.

Question Info

Course: CHM 102
Topic: Radioactivity
Type: Objective
Difficulty: Not specified
Question 1875 of 1999
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