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Chapter 29: Problem 51

Irène and Jean Frédéric Joliot-Curie, in an experiment that led to the 1935 Nobel Prize in chemistry, bombarded aluminum \(_{13}^{27}\) Al with \(\alpha\) particles to form a highly unstable isotope of phosphorus, $_{15}^{31} \mathrm{P}$. The phosphorus immediately decayed into another isotope of phosphorus, \(_{15}^{30} \mathrm{P}\), plus another product. Write out these reactions, identifying the other product.

Short Answer

Expert verified
Answer: The other product formed in the second reaction is a neutron, represented as \(_0^1\)n.

Step by step solution

01

Write the first nuclear reaction

In the first reaction, aluminum \(_{13}^{27}\)Al gets bombarded by α-particles, which have two protons and two neutrons, represented as \(_2^4\)He. We need to find the resulting product when \(_{13}^{27}\)Al reacts with \(_2^4\)He. We can do this by adding the atomic and mass numbers for aluminum and the α-particle. The atomic number (Z) for the product P is the sum of the atomic numbers for aluminum and α-particle: \(Z_{Al} + Z_{He} = 13 + 2 = 15\). Similarly, the mass number (A) for the product P is the sum of the mass numbers for aluminum and α-particle: \(A_{Al} + A_{He} = 27 + 4 = 31\). So, the first reaction is: \(_{13}^{27}\)Al + \(_2^4\)He → \(_{15}^{31}\)P
02

Write the second nuclear reaction

In the second reaction, the unstable phosphorus isotope \(_{15}^{31}\)P decays into a stable isotope, \(_{15}^{30}\)P. To find the other product, we can use the conservation of mass and atomic numbers. The atomic number (Z) and mass number (A) for the other product X are: \(Z_X = Z_{first~ product} - Z_{second ~product} = 15 - 15 = 0\) \(A_X = A_{first~ product} - A_{second ~product} = 31 - 30 = 1\) The product with atomic number 0 and mass number 1 is a neutron, represented as \(_0^1\)n. So, the second reaction is: \(_{15}^{31}\)P → \(_{15}^{30}\)P + \(_0^1\)n

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