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Would you please check my answers. I think it should be all okay but if not, could you please help me out then?

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C. Bismuth (Bi) is an element that exists as multiple isotopes, some of which are radioactive. i. Complete Table 1 for two of the neutral isotopes of bismuth. Table 1 The number of atomic particles in two neutral isotopes of bismuth. Isotope Protons Neutrons Electrons Bi-209 23 Bi-213 83 130 83 Bi-213 is used in cancer therapy. Figure 2 shows a plot of the radioactive decay of Bi-213. 25 000 20 000 15 000 number of Bi-213 nuclei 10 000 5 000 O O 20 40 60 80 100 120 140 160 180 200 time/minutes Figure 2 A plot of the radioactive decay of Bi-213 Use Figure 2 to work out the half-life of Bi-213 by considering two consecutive half-lives for the isotope.\fAnswer: 90 minutes Explanation: after one half-life, N1 : = No X after two half-lives N2 = N1 x7 = (Nox=) X- = No (-)2 With value: 1 (2 x2) 2 = NIH = 2 x 45 minutes = 90 minutes The two half-live isotopes of Bi-213 nuclei is 90 minutes.Give the definition of the following terms: - Alpha particles are positively charged nuclear particles (the nucleus of a : 4 2 helium atom EHE)' They consist of two protons and two neutrons bound - Beta particles (p) are subatomic particles with a small mass of around 1/2000th the mass of a neutron. They can have a positive or negative charge. High-energy electrons are the most common negative beta particles, while positively charged beta particles are called positrons and are usually generated during the decay of synthetic radioisotopes. Gamma rays (y) are high-energy electromagnetic radiation that can penetrate most materials, including the human body. They are the most dangerous and require high-density materials like concrete or lead to stop their penetration. If the dose rate from a sample of Bi-213 was found to be 5.4 x 1075 mSv per hour at 1.6 metres, calculate the dose rate at 4.8 metres. Show your calculations. Give your answer to two significant figures. Answer: 6.0 x 106 mSv/hour Calculation: D1 = 5.4 x 10-5 mSv r1 = 1.6 m D2 =? r2 = 4.8 m Re-arranging equation D2 = D1 x Inserting values, D2: 5.4 x 10-5 mSv hour x x (1.6 m) 4.8 m 5.4 x 10-5 mSv hour x () 5.4 x 10-5 mSv = hour x - = 6.0 x 10-6 mSv/hour Therefore, the dose rate of Bi-213 at 4.8 metres is 6.0 x 106 mSv/hour to two significant figures