When the isotope actinium-227 (227 89 Ac) undergoes alpha decay, what is the resulting nucleus?

• A. 225 88 Ra
• B. 223 88 Ra
• C. 225 87 Fr
• D. 223 87 Fr

Answer: (D)

Verified step by step guidance

1. Recall that alpha decay involves the emission of an alpha particle, which is a helium nucleus consisting of 2 protons and 2 neutrons. This means the original nucleus loses 2 protons and 2 neutrons.
2. Write down the original isotope: actinium-227, which has an atomic number (number of protons) of 89 and a mass number (total protons + neutrons) of 227, represented as \(^{227}_{89}Ac\).
3. Subtract 2 from the atomic number to account for the loss of 2 protons: \$89 - 2 = 87$. This gives the atomic number of the new element.
4. Subtract 4 from the mass number to account for the loss of 2 protons and 2 neutrons: \$227 - 4 = 223$. This gives the mass number of the new nucleus.
5. Identify the element with atomic number 87, which is francium (Fr), so the resulting nucleus is \(^{223}_{87}Fr\).