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Write nuclear equations for beta decay and radioactive decay, and discuss their effects and graphing.

Physics

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6. (a)

  1. Balance the mass numbers (superscripts).

    131=131+x  ⟹  x=0131 = 131 + x \implies x = 0
  2. Balance the atomic numbers (subscripts).

    53=54+y  ⟹  y=−153 = 54 + y \implies y = -1
  3. Identify the particle.

    A particle with mass number 0 and atomic number -1 is a beta particle (electron).
    The complete nuclear equation is:

    53131I⟶54131Xe+−10e^{131}_{53}I \longrightarrow ^{131}_{54}Xe + \boxed{^{0}_{-1}e}

    6. (b)
    Assumption: The initial mass of 53131I^{131}_{53}I is 50 grams.

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

    n=total timehalf-life=40 days8 days=5n = \frac{\text{total time}}{\text{half-life}} = \frac{40 \text{ days}}{8 \text{ days}} = 5
  5. Calculate the remaining mass (MtM_t).

    Mt=M0×(12)nM_t = M_0 \times \left(\frac{1}{2}\right)^n Mt=50 g×(12)5M_t = 50 \text{ g} \times \left(\frac{1}{2}\right)^5 Mt=50 g×132M_t = 50 \text{ g} \times \frac{1}{32} Mt=5032 g=2516 g=1.5625 gM_t = \frac{50}{32} \text{ g} = \frac{25}{16} \text{ g} = 1.5625 \text{ g}

    The mass of 53131I^{131}_{53}I remaining is 1.5625 g\boxed{1.5625 \text{ g}}.

    6. (c)
    One harmful effect of radioisotopes is increased risk of cancer.

    7.
    To plot a graph of the mass of iodine-131 against time, you would use the following data points:

    • At time = 0 days, mass = 120 g
    • At time = 8 days (1 half-life), mass = 120 g×12=60 g120 \text{ g} \times \frac{1}{2} = 60 \text{ g}
    • At time = 16 days (2 half-lives), mass = 60 g×12=30 g60 \text{ g} \times \frac{1}{2} = 30 \text{ g}
    • At time = 24 days (3 half-lives), mass = 30 g×12=15 g30 \text{ g} \times \frac{1}{2} = 15 \text{ g}
    • At time = 32 days (4 half-lives), mass = 15 g×12=7.5 g15 \text{ g} \times \frac{1}{2} = 7.5 \text{ g} Plot these points on a graph with time on the x-axis and mass on the y-axis, then draw a smooth curve connecting them.

    8.
    Alpha, beta, and gamma radiations can be distinguished by their behavior in an electric or magnetic field:

    • Alpha (α\alpha) particles are positively charged and are deflected towards the negative plate in an electric field. They have the largest mass and least penetrating power.
    • Beta (β\beta) particles are negatively charged and are deflected towards the positive plate in an electric field. They have a much smaller mass than alpha particles and are more penetrating.
    • Gamma (γ\gamma) rays are electromagnetic radiation with no charge and no mass, so they are not deflected by electric or magnetic fields. They have the highest penetrating power.

    9. (a)

    • A neutron is a subatomic particle with no electric charge and a mass slightly greater than a proton.
    • A proton is a subatomic particle with a positive electric charge (+1e) and a mass slightly less than a neutron.

    9. (b)
    A radioactive substance is a substance that undergoes spontaneous nuclear decay, emitting radiation (such as alpha, beta, or gamma particles/rays) to transform into a more stable nucleus.

    9. (c)
    Two dangers associated with radioactive substances in the environment are:

    • Contamination of water and food sources, leading to internal exposure when ingested.
    • Damage to living organisms and ecosystems, causing cell damage, genetic mutations, and increased risk of cancer in plants, animals, and humans.

    9. (d) (i)
    The nuclear equation is: 12D+13T⟶ZAY+01n^{2}_{1}D + ^{3}_{1}T \longrightarrow ^{A}_{Z}Y + ^{1}_{0}n
    (I) Mass of Y:

  6. Balance the mass numbers (superscripts).

    2+3=A+12 + 3 = A + 1 5=A+15 = A + 1 A=4A = 4

    The mass of Y is 4\boxed{4}.

    (II) Number of Y:

  7. Balance the atomic numbers (subscripts).

    1+1=Z+01 + 1 = Z + 0 2=Z2 = Z

    The atomic number of Y is 2\boxed{2}.

    9. (d) (ii)
    The type of reaction undergone by the isotopes of hydrogen is nuclear fusion.

    9. (e) (i)
    The half-life of a radioactive substance is the time taken for half of the radioactive nuclei in a given sample to decay.

    9. (e) (ii)

  8. Determine the number of half-lives (nn).

    Mt=M0×(12)nM_t = M_0 \times \left(\frac{1}{2}\right)^n 9 g=288 g×(12)n9 \text{ g} = 288 \text{ g} \times \left(\frac{1}{2}\right)^n 9288=(12)n\frac{9}{288} = \left(\frac{1}{2}\right)^n 132=(12)n\frac{1}{32} = \left(\frac{1}{2}\right)^n

    Since 25=322^5 = 32, then (12)5=132\left(\frac{1}{2}\right)^5 = \frac{1}{32}.
    Therefore, n=5n = 5.

  9. Calculate the half-life (T1/2T_{1/2}).

    n=total timeT1/2n = \frac{\text{total time}}{T_{1/2}} 5=40 daysT1/25 = \frac{40 \text{ days}}{T_{1/2}} T1/2=40 days5=8 daysT_{1/2} = \frac{40 \text{ days}}{5} = 8 \text{ days}

    The half-life of the radioactive substance is 8 days\boxed{8 \text{ days}}.

    10. (a)

  10. Balance the mass numbers (superscripts).

    226=222+x  ⟹  x=4226 = 222 + x \implies x = 4
  11. Balance the atomic numbers (subscripts).

    88=86+y  ⟹  y=288 = 86 + y \implies y = 2
  12. Identify the particle.

    A particle with mass number 4 and atomic number 2 is an alpha particle.
    The complete nuclear reaction is:

    88226Q⟶86222P+24α^{226}_{88}Q \longrightarrow ^{222}_{86}P + \boxed{^{4}_{2}\alpha}

    10. (b)
    Two uses of radioisotopes in health are:

    • Medical imaging for diagnosis (e.g., using technetium-99m for bone scans or iodine-131 for thyroid scans).
    • Radiotherapy for cancer treatment (e.g., using cobalt-60 or iridium-192 to destroy cancerous cells).
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