Bismuth-210 is beta emitter with a half-life of 5.0 days. Part A: If a sample co

Question

Bismuth-210 is beta emitter with a half-life of 5.0 days.

Part A:

If a sample contains 1.2 g of Bi-210 (atomic mass = 209.984105 amu), how many beta emissions would occur in 13.0 days?

Express your answer using two significant figures


? beta decays


Part B:

If a person's body intercepts 5.8% of those emissions, to what dose of radiation (in Ci) is the person exposed?


Express your answer using two significant figures

Explanation / Answer

Part - A Given: t1/2=5 days --> k=.693/5 = .1386 days^-1 t=13.5 days Equation: ln(final) = -.1386(13) + ln(1.2g) final = .1847 initial-final = 1.2 - .1847 = 1.0153g * mol/209.98gBi = 4.858e-3 mol Bi used 4.858e-3 mol * 6.022e23 beta emissions/mol = 2.9255e21 beta emissions Part - B As for the second part: 13 days = 1123200 seconds (2.9255e21 beta emissions*.055)/1123200 seconds = 1.433 * e14 emissions/second Since 3.7e10 emissions/second = 1 Ci, (1.433e14 emissions/second) / (3.7e10 emissions/second) = 3873 Ci

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.