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Calculate the initial mass of the isotope. 10. a) Name the radioactive process displayed above. Give a reason for your answer. b) Suggest one application of t

Physics

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9. Calculate the initial mass of the isotope.

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

    The total time is 1000 days, and the half-life is 25 days.

    n=Total timeHalf-life=1000 days25 days=40n = \frac{\text{Total time}}{\text{Half-life}} = \frac{1000 \text{ days}}{25 \text{ days}} = 40

    The substance undergoes 40 half-lives.

  2. Use the formula for radioactive decay to find the initial mass (N0N_0).

    The formula is Nt=N0(12)nN_t = N_0 \left(\frac{1}{2}\right)^n, where NtN_t is the final mass, N0N_0 is the initial mass, and nn is the number of half-lives.
    We are given Nt=50 gN_t = 50 \text{ g} and n=40n = 40.

    50 g=N0(12)4050 \text{ g} = N_0 \left(\frac{1}{2}\right)^{40}
  3. Solve for N0N_0.

    N0=50 g(12)40=50 g×240N_0 = \frac{50 \text{ g}}{\left(\frac{1}{2}\right)^{40}} = 50 \text{ g} \times 2^{40} N0=50 g×1.0995×1012N_0 = 50 \text{ g} \times 1.0995 \times 10^{12} N0=5.4975×1013 gN_0 = 5.4975 \times 10^{13} \text{ g}

    The initial mass of the isotope was approximately 5.50×10135.50 \times 10^{13} grams.

    The initial mass of the isotope is 5.50×1013 g\boxed{\text{5.50} \times 10^{13} \text{ g}}.

    10.

    a) Name the radioactive process displayed above. Give a reason for your answer.

    The process is Nuclear Fission.
    Reason: A heavy nucleus (23592U\frac{235}{92}U) absorbs a neutron and splits into two lighter nuclei (14256Ba\frac{142}{56}Ba and 9136Kr\frac{91}{36}Kr), releasing energy and additional neutrons.

    b) Suggest one application of this process.

    One application is electricity generation in nuclear power plants.

    11.

    a) Complete the nuclear equation below.

    13153I⟶13154Xe+0−1e\frac{131}{53}I \longrightarrow \frac{131}{54}Xe + \boxed{\frac{0}{-1}e}

    This is a beta decay, where a neutron in the nucleus converts into a proton, emitting an electron (beta particle).

    b) The half-life of 13153I\frac{131}{53}I is 8 days. Determine the mass of 13153I\frac{131}{53}I remaining if 50g decayed for 40 days.

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

    The total time is 40 days, and the half-life is 8 days.

    n=Total timeHalf-life=40 days8 days=5n = \frac{\text{Total time}}{\text{Half-life}} = \frac{40 \text{ days}}{8 \text{ days}} = 5

    The substance undergoes 5 half-lives.

  5. Determine the initial mass.

    The question states "if 50g decayed for 40 days", implying the initial mass was 50g.
    Initial mass (N0N_0) = 50 g.

  6. Calculate the remaining mass (NtN_t).

    Nt=N0(12)nN_t = N_0 \left(\frac{1}{2}\right)^n Nt=50 g×(12)5N_t = 50 \text{ g} \times \left(\frac{1}{2}\right)^5 Nt=50 g×132N_t = 50 \text{ g} \times \frac{1}{32} Nt=5032 g=2516 g=1.5625 gN_t = \frac{50}{32} \text{ g} = \frac{25}{16} \text{ g} = 1.5625 \text{ g}

    The mass of 13153I\frac{131}{53}I remaining is 1.5625 g\boxed{\text{1.5625 g}}.

    c) Give one harmful effect of radioisotopes.

    One harmful effect is causing cancer (e.g., leukemia, thyroid cancer) due to damage to DNA and cells.

    12.

    a) Identify radiation W.

    Radiation W passes through paper and aluminium foil but is stopped by a lead block, and it is undeflected by the electric field. This indicates it is Gamma radiation (γ\gamma).

    b) Which of the radiations would cause most danger to human tissues? Explain your answer.

    Gamma radiation (W) would cause the most danger.
    Explanation: Gamma radiation has very high penetrating power, meaning it can pass through the body and damage internal organs and tissues, leading to widespread cellular damage and increased risk of cancer.

    c) Explain why it is not safe to store radioactive substances in containers made of iron sheets.

    It is not safe because iron sheets are not dense or thick enough to effectively block all types of radiation, especially gamma rays. Gamma radiation can easily penetrate iron sheets, allowing harmful radiation to escape and pose a health hazard to people nearby.

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