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MCAT (Medical College Admission Test) Interview Question:
When 47Be undergoes radioactive decay by electron capture (a form of ß+ decay), the resulting nucleus is:
Submitted by: Administrator1. 36Li
2. 37Li
3. 47Be
4. 48Be
Answer: B
Explanation: In radioactive decay, the sum of the mass numbers A and atomic numbers Z, before and after decay, must balance. The numbers for beryllium undergoing positron decay are: mass (7 = 7 + 0) and atomic (4 = 3 + 1). The resulting nucleus is 73Li. Thus, answer choice B is the best answer.
Submitted by: Administrator
2. 37Li
3. 47Be
4. 48Be
Answer: B
Explanation: In radioactive decay, the sum of the mass numbers A and atomic numbers Z, before and after decay, must balance. The numbers for beryllium undergoing positron decay are: mass (7 = 7 + 0) and atomic (4 = 3 + 1). The resulting nucleus is 73Li. Thus, answer choice B is the best answer.
Submitted by: Administrator
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