When Mendelevium-255 () is bombarded with an -particle, it undergoes an artificial transmutation tha β Nuclear Chemistry and Radioactivity Chemistry Question
Question
When Mendelevium-255 (${}_{101}^{255}Md$) is bombarded with an $\alpha$ -particle, it undergoes an artificial transmutation that emits two neutrons. What is the stable or intermediate product nuclide formed in this reaction?
Answer: C
π‘ Solution & Explanation
The nuclear reaction can be written as: ${}_{101}^{255}Md + {}_{2}^{4}He \rightarrow {}_{Z}^{A}X + 2 {}_{0}^{1}n$. Balancing the mass numbers: $255 + 4 = A + 2(1) \Rightarrow A = 257$. Balancing the atomic numbers: $101 + 2 = Z + 2(0) \Rightarrow Z = 103$. The element with atomic number $103$ is Lawrencium ($Lr$). Thus, the product is ${}_{103}^{257}Lr$.
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