During the first artificially induced nuclear transmutation carried out by Irene and Frederick Jolio — Nuclear Chemistry and Radioactivity Chemistry Question
Question
During the first artificially induced nuclear transmutation carried out by Irene and Frederick Joliot-Curie, Aluminum-27 (${}_{13}^{27}Al$) was bombarded with $\alpha$ -particles (${}_{2}^{4}He$) to form a radioactive isotope of Phosphorus (${}_{15}^{30}P$) alongside an emitted particle $X$. What is the mass number of this emitted particle $X$?
Answer: 1
💡 Solution & Explanation
The complete balanced nuclear reaction is: ${}_{13}^{27}Al + {}_{2}^{4}He \rightarrow {}_{15}^{30}P + {}_{Z}^{A}X$. Balancing the mass numbers (superscripts): $27 + 4 = 30 + A \Rightarrow A = 1$. Balancing the atomic numbers (subscripts): $13 + 2 = 15 + Z \Rightarrow Z = 0$. The particle with mass number 1 and atomic number 0 is a neutron (${}_{0}^{1}n$). Its mass number is 1.
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