Question

In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bombarded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process.

Short answer

The element produced when molybdenum-96 was bombarded with deuterium ions is technetium (Tc), specifically \(^{98}_{43}\text{Tc}\). The nuclear equation representing this process is: \[^{96}_{42}\text{Mo} + ^{2}_{1}\text{H} \rightarrow ^{98}_{43}\text{Tc} + ^{0}_{0}\gamma\]

Step-by-step solution

Identify the reactants

In this problem, we have two reactants: 1. Molybdenum-96 (Mo-96), which is given in the problem. The atomic number of molybdenum is 42, so Mo-96 can be represented as \(^{96}_{42}\text{Mo}\). 2. Deuterium ions, which are hydrogen ions with one proton and one neutron, and can be represented as \(^{2}_{1}\text{H}\).

Determine the product element

When a molybdenum target is bombarded with deuterium ions, a new nucleus is formed. To find the new element, we need to add the atomic numbers and mass numbers of the reactants. 1. The atomic number of the new element would be the sum of the atomic numbers of molybdenum and deuterium: 42 + 1 = 43. 2. The mass number of the new element would be the sum of the mass numbers of molybdenum and deuterium: 96 + 2 = 98. By looking up the element with atomic number 43 in the periodic table, we find that the new element is technetium (Tc). So, the new nucleus will be \(^{98}_{43}\text{Tc}\).

Write the nuclear equation

Now that we know the reactants and the product nucleus, we can write the nuclear equation for this process: \[^{96}_{42}\text{Mo} + ^{2}_{1}\text{H} \rightarrow ^{98}_{43}\text{Tc} + X\] The reaction might produce additional particles, represented by 'X' in the equation. To find out what particle 'X' is, we can use the conservation of mass and atomic numbers. 1. The total initial atomic number on the left side is 42 + 1 = 43. The total atomic number on the right side is 43 + Z, where Z is the atomic number of particle 'X'. Thus, Z = 0. 2. The total initial mass number on the left side is 96 + 2 = 98. The total mass number on the right side is 98 + A, where A is the mass number of particle 'X'. Thus, A = 0. Particle 'X' has an atomic number of 0 and a mass number of 0, which means it is a gamma ray (γ).

Complete the nuclear equation

Now that we have identified particle 'X', we can complete the nuclear equation: \[^{96}_{42}\text{Mo} + ^{2}_{1}\text{H} \rightarrow ^{98}_{43}\text{Tc} + ^{0}_{0}\gamma\] This is the nuclear equation representing the process of producing technetium from bombarding a molybdenum-96 target with deuterium ions.