Question

which isotope is produced if radium - 226 undergoes alpha decay?

• radon - 222
• francium - 230
• actinium - 226
• thorium - 224

a 10.0 g sample of strontium - 90 decays for 87 years leaving 1.25 grams undecayed. what is the isotopes half - life?

• 29 years
• 1 year
• 43.5 years
• 5.6 years

Explanation:

Step1: Recall alpha - decay rule

In alpha - decay, an atom loses 2 protons and 2 neutrons. Radium - 226 has an atomic number of 88. After alpha - decay, the mass number decreases by 4 and the atomic number decreases by 2. The new mass number is \(226 - 4=222\) and the new atomic number is \(88 - 2 = 86\). The element with atomic number 86 is radon. So the isotope produced is radon - 222.

Step2: Use radioactive decay formula

The radioactive decay formula is \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), where \(N\) is the remaining amount, \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(T_{1/2}\) is the half - life. We know that \(N_0 = 10.0\ g\), \(N = 1.25\ g\), and \(t = 87\) years. Substitute these values into the formula: \(1.25=10.0(\frac{1}{2})^{\frac{87}{T_{1/2}}}\). First, divide both sides by 10.0: \(\frac{1.25}{10.0}=(\frac{1}{2})^{\frac{87}{T_{1/2}}}\), so \(0.125 = (\frac{1}{2})^{\frac{87}{T_{1/2}}}\). Since \(0.125=\frac{1}{8}=(\frac{1}{2})^3\), then \(\frac{87}{T_{1/2}} = 3\). Solve for \(T_{1/2}\): \(T_{1/2}=\frac{87}{3}=29\) years.

Answer:

1. Radon - 222
2. 29 years