Question

uranium - 238 decays to lead - 206 through a series of nuclear reactions. only α - particles and β - particles are emitted. how many α - particles are emitted?
7
8
6
10

which isotope is produced if plutonium - 242 undergoes alpha decay?
uranium - 238
uranium - 236
americium - 242
plutonium - 243

beta emission is associated with

Explanation:

1. First question:

• Explanation:
• In nuclear - decay, when uranium - 238 decays to lead - 206 through a series of nuclear reactions, the change in mass number (\(\Delta A\)) and atomic number (\(\Delta Z\)) is related to the number of \(\alpha\) and \(\beta\) particles emitted. An \(\alpha\) - particle has a mass number of 4 and an atomic number of 2 (\(_{2}^{4}He\)), and a \(\beta\) - particle has a mass number of 0 and an atomic number of - 1 (\(_{ - 1}^{0}e\)).
• The change in mass number \(\Delta A=238 - 206 = 32\). Since the mass number of an \(\alpha\) - particle is 4, the number of \(\alpha\) - particles emitted \(n_{\alpha}=\frac{238 - 206}{4}=8\).
• The change in atomic number for uranium (\(Z = 92\)) to lead (\(Z = 82\)) is \(\Delta Z=92 - 82 = 10\). The contribution of \(\alpha\) - particles to the change in atomic number is \(2n_{\alpha}=2\times8 = 16\). Let the number of \(\beta\) - particles be \(n_{\beta}\). Then, \(92-2n_{\alpha}+n_{\beta}=82\). Substituting \(n_{\alpha}=8\) into the equation: \(92-16 + n_{\beta}=82\), so \(n_{\beta}=6\).
• Answer: 6

1. Second question:

• Explanation:
• In alpha - decay, the mass number of the parent nucleus decreases by 4 and the atomic number decreases by 2. Plutonium has an atomic number \(Z = 94\).
• For an alpha - decay of plutonium - 242 (\(_{94}^{242}Pu\)), the mass number of the daughter nucleus \(A=242 - 4=238\) and the atomic number \(Z = 94 - 2 = 92\). The element with atomic number 92 is uranium. So the daughter nucleus is uranium - 238 (\(_{92}^{238}U\)).
• Answer: Uranium - 238

1. Third question:

• Explanation:
• Beta - emission is associated with the conversion of a neutron into a proton and an electron (beta - particle). In beta - decay, a neutron (\(_{0}^{1}n\)) changes to a proton (\(_{1}^{1}p\)) and an electron (\(_{ - 1}^{0}e\)) (beta - particle) and an antineutrino (\(\bar{

u}\)). This process increases the atomic number of the nucleus by 1 while the mass number remains the same.

• Answer: The conversion of a neutron into a proton and an electron.

Step1: Analyze first - decay problem

For the uranium - 238 to lead - 206 decay, find number of \(\alpha\) - particles from mass - number change. \(\frac{238 - 206}{4}=8\).

Step2: Find number of \(\beta\) - particles

Use atomic - number change equation \(92-2\times8 + n_{\beta}=82\), solve for \(n_{\beta}\), \(n_{\beta}=6\).

Step3: Analyze plutonium alpha - decay

For plutonium - 242 alpha - decay, calculate mass number \(242 - 4 = 238\) and atomic number \(94 - 2 = 92\), which gives uranium - 238.

Step4: Explain beta - emission

Beta - emission is neutron to proton and electron conversion.

Answer:

6
Uranium - 238
The conversion of a neutron into a proton and an electron.