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Distinguish between natural and induced radioactivity. Calculate the half-life of a radioactive substance given initial and final masses and time elapsed.

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c) i)
Natural radioactivity is the spontaneous disintegration of unstable atomic nuclei found in nature, without any external intervention.
Induced radioactivity is the artificial creation of unstable atomic nuclei by bombarding stable nuclei with high-energy particles (like neutrons or protons), causing them to become radioactive.

c) ii)

  1. Identify the given values.

    Initial mass, N0=0.600gN_0 = 0.600\text{g}
    Final mass, Nt=0.395gN_t = 0.395\text{g}
    Time elapsed, t=10t = 10 years

  2. Use the radioactive decay formula.

    The formula relating the remaining mass to the initial mass and half-life is:

    Nt=N0(12)t/TN_t = N_0 \left(\frac{1}{2}\right)^{t/T}

    where TT is the half-life.

  3. Substitute the given values into the formula.

    0.395g=0.600g(12)10/T0.395\text{g} = 0.600\text{g} \left(\frac{1}{2}\right)^{10/T}

  4. Isolate the exponential term.

    0.3950.600=(12)10/T\frac{0.395}{0.600} = \left(\frac{1}{2}\right)^{10/T} 0.658333...=(12)10/T0.658333... = \left(\frac{1}{2}\right)^{10/T}

  5. Take the natural logarithm of both sides.

    ln⁡(0.658333...)=ln⁡((12)10/T)\ln(0.658333...) = \ln\left(\left(\frac{1}{2}\right)^{10/T}\right) ln⁡(0.658333...)=10Tln⁡(12)\ln(0.658333...) = \frac{10}{T} \ln\left(\frac{1}{2}\right)

  6. Solve for TT.

    We know that ln⁡(12)=−ln⁡(2)\ln\left(\frac{1}{2}\right) = -\ln(2).

    ln⁡(0.658333...)=10T(−ln⁡(2))\ln(0.658333...) = \frac{10}{T} (-\ln(2)) T=10×(−ln⁡(2))ln⁡(0.658333...)T = \frac{10 \times (-\ln(2))}{\ln(0.658333...)} T=−10×0.693147−0.41800T = \frac{-10 \times 0.693147}{-0.41800} T=6.931470.41800T = \frac{6.93147}{0.41800} T≈16.582 yearsT \approx 16.582 \text{ years}

    The half-life of 90Sr^{90}\text{Sr} is approximately 16.6 years.

    16.6 years\boxed{16.6 \text{ years}}
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