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Determine the original mass of a radioactive substance given its half-life and remaining mass, list two medical uses of radioactivity, and solve nuclear equations.

Physics

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here are the answers to the questions:

11. (a)

  1. Calculate the number of half-lives (nn).

    n=total timehalf-life=1900 years380 years=5n = \frac{\text{total time}}{\text{half-life}} = \frac{1900 \text{ years}}{380 \text{ years}} = 5
  2. Use the formula for radioactive decay to find the original mass (M0M_0).

    Mt=M0×(12)nM_t = M_0 \times \left(\frac{1}{2}\right)^n

    Given Mt=15 kgM_t = 15 \text{ kg} and n=5n = 5:

    15 kg=M0×(12)515 \text{ kg} = M_0 \times \left(\frac{1}{2}\right)^5 15 kg=M0×13215 \text{ kg} = M_0 \times \frac{1}{32}
  3. Solve for M0M_0.

    M0=15 kg×32M_0 = 15 \text{ kg} \times 32 M0=480 kgM_0 = 480 \text{ kg}

    The original mass of the radioactive substance is 480 kg\boxed{480 \text{ kg}}.

    11. (b)
    Two uses of radioactivity in medicine are:

    • Sterilization of medical equipment (e.g., using gamma radiation).
    • Diagnosis of diseases (e.g., using radioisotopes as tracers for imaging organs).

    12.
    The nuclear equation is:

    92239U⟶XYTh+224α+2−10β^{239}_{92}U \longrightarrow ^{Y}_{X}Th + 2^{4}_{2}\alpha + 2^{0}_{-1}\beta
  4. Balance the mass numbers (superscripts) to find Y.

    239=Y+(2×4)+(2×0)239 = Y + (2 \times 4) + (2 \times 0) 239=Y+8+0239 = Y + 8 + 0 239=Y+8239 = Y + 8 Y=239−8=231Y = 239 - 8 = 231
  5. Balance the atomic numbers (subscripts) to find X.

    92=X+(2×2)+(2×−1)92 = X + (2 \times 2) + (2 \times -1) 92=X+4−292 = X + 4 - 2 92=X+292 = X + 2 X=92−2=90X = 92 - 2 = 90

    The values are X=90\boxed{X = 90} and Y=231\boxed{Y = 231}.

    13. (a) (i)
    To draw a graph of disintegration rate against time, plot the following points with Time (hours) on the x-axis and Rate of disintegration (counts s⁻¹) on the y-axis:

    • (0, 730)
    • (5, 570)
    • (10, 455)
    • (15, 365)
    • (20, 292)
    • (25, 232) Draw a smooth curve connecting these points.

    13. (a) (ii)

  6. Determine the initial disintegration rate from the table.

    Initial rate at time = 0 hours is 730 counts s⁻¹.

  7. Calculate half of the initial rate.

    Half of 730 counts s⁻¹ is 730/2=365730 / 2 = 365 counts s⁻¹.

  8. Find the time at which the disintegration rate is 365 counts s⁻¹ from the table.

    From the table, at 15 hours, the rate of disintegration is 365 counts s⁻¹.
    The half-life of bismuth is 15 hours\boxed{15 \text{ hours}}.

    13. (a) (iii)
    If half the amount of sample of bismuth were used, the initial disintegration rate would be halved. The curve would start at 365 counts s⁻¹ (instead of 730 counts s⁻¹) and follow the same decay pattern, but all disintegration rate values would be half of the original values. The half-life would remain the same.

    13. (b)
    One application of radioactivity in:

    • (i) Medicine: Radiotherapy for cancer treatment.
    • (ii) Agriculture: Pest control through sterilization of male insects.
    • (iii) Tracers: Detecting leaks in underground pipes.
    • (iv) Nuclear power station: Generating electricity through nuclear fission.

    13. (c)
    Two dangers associated with radioactivity are:

    • Genetic mutations and birth defects in offspring due to damage to reproductive cells.
    • Radiation sickness from high doses, leading to nausea, vomiting, hair loss, and potentially death.

    14. (a)
    The two charged particles emitted by a radioactive isotope are:

    • Alpha particle: 24α\boxed{^{4}_{2}\alpha} or 24He\boxed{^{4}_{2}He} (positive charge)
    • Beta particle: −10β\boxed{^{0}_{-1}\beta} or −10e\boxed{^{0}_{-1}e} (negative charge)

    14. (b)
    The nuclear equation is:

    82210Pb⟶ZAX+2−10β^{210}_{82}Pb \longrightarrow ^{A}_{Z}X + 2^{0}_{-1}\beta
  9. Balance the mass numbers (superscripts).

    210=A+(2×0)210 = A + (2 \times 0) 210=A+0210 = A + 0 A=210A = 210
  10. Balance the atomic numbers (subscripts).

    82=Z+(2×−1)82 = Z + (2 \times -1) 82=Z−282 = Z - 2 Z=82+2=84Z = 82 + 2 = 84

    The mass number of the resulting nuclide is 210\boxed{210} and the atomic number is 84\boxed{84}.

    15.
    The diagram in Figure 6 is not provided in the image, so this question cannot be answered.

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