This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the answers to the questions: 6. (a) Step 1: Balance the mass numbers (superscripts). 131 = 131 + x x = 0 Step 2: Balance the atomic numbers (subscripts). 53 = 54 + y y = -1 Step 3: Identify the particle. A particle with mass number 0 and atomic number -1 is a beta particle (electron). The complete nuclear equation is: ^131_53I ^131_54Xe + ^0_-1e 6. (b) Assumption: The initial mass of ^131_53I is 50 grams. Step 1: Calculate the number of half-lives (n). n = total timehalf-life = 40 days8 days = 5 Step 2: Calculate the remaining mass (M_t). M_t = M_0 × ((1)/(2))^n M_t = 50 g × ((1)/(2))^5 M_t = 50 g × (1)/(32) M_t = (50)/(32) g = (25)/(16) g = 1.5625 g The mass of ^131_53I remaining is 1.5625 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 × (1)/(2) = 60 g • At time = 16 days (2 half-lives), mass = 60 g × (1)/(2) = 30 g • At time = 24 days (3 half-lives), mass = 30 g × (1)/(2) = 15 g • At time = 32 days (4 half-lives), mass = 15 g × (1)/(2) = 7.5 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 () 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 () 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 () 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: ^2_1D + ^3_1T ^A_ZY + ^1_0n (I) Mass of Y: Step 1: Balance the mass numbers (superscripts). 2 + 3 = A + 1 5 = A + 1 A = 4 The mass of Y is 4. (II) Number of Y: Step 1: Balance the atomic numbers (subscripts). 1 + 1 = Z + 0 2 = Z The atomic number of Y is 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) Step 1: Determine the number of half-lives (n). M_t = M_0 × ((1)/(2))^n 9 g = 288 g × ((1)/(2))^n (9)/(288) = ((1)/(2))^n (1)/(32) = ((1)/(2))^n Since 2^5 = 32, then ((1)/(2))^5 = (1)/(32). Therefore, n = 5. Step 2: Calculate the half-life (T_1/2). n = total timeT_1/2 5 = 40 daysT_1/2 T_1/2 = 40 days5 = 8 days The half-life of the radioactive substance is 8 days. 10. (a) Step 1: Balance the mass numbers (superscripts). 226 = 222 + x x = 4 Step 2: Balance the atomic numbers (subscripts). 88 = 86 + y y = 2 Step 3: Identify the particle. A particle with mass number 4 and atomic number 2 is an alpha particle. The complete nuclear reaction is: ^226_88Q ^222_86P + ^4_2 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). 3 done, 2 left today. You're making progress.