Hello,

Could someone check if my answers are correct, please? (I like to think there are, but if not?)

Thank you

If the dose rate from a sample of Bi-213 was found to be 5.4 x 10-5 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 in two significant figures Explanation: Using the inverse square law, with this equation: h1 x d12 = 12 x d12 Re-arranging equation: hxd- 12 = Adding the numbers so /2: (5.4x10-5 mSv/hr) x (1.6 m)2 (4.8 m)2 = max m2 = 1 (5.4x10 5 mSv/hr) x 1.6 m 4.8 m = 6 x 10-6 mSv/hr Therefore, the dose rate of Bi-213 at 4.8 metres is 6.0 x 10-6 mSv/hour in SI units and two significant figures.Calculate the ratio of the temperature at the centre of the Sun compared to the temperature of the photosphere. Express your final answer in the fully factorised form Teentre : Tphotosphere = 1 : X, Where x is a number that you should determine to an appropriate number of significant figures and write using scientific notation. The sun core temperature is approximately 15 millions Kelvin The sun photosphere varies but is approximately 5813 Kelvin Tcentre Tphotosphere _ 15000000 K ~ 5813K = 2580.7 K Therefore, Tcentre : I photosphere has a ratio of 1 : 2.581 x 103 Kelvin, in scientific notation and in four significant figures. The red supergiant Betelgeuse is estimated to have a luminosity of 4.82 x 1031 W. How much energy does it radiate in 1.00 day? Give your answer using scientific notation, the SI unit for energy and an appropriate number of significant figures. The energy radiated by a star is given by this formula.: E = P xt Where: E is the energy, P is the power (or luminosity), and t is the time. Knowing: P = 4.82 x 1031 W (Watts, joules per second) t = 1.00 day = 24 h d-1 x 60 min h-l = 1440 min h-1 x 60 s min-1 = 86400 s min-1 or = 8.64 x 10+ s (in scientific notation) Inserting the values so E: = (4.82 x 1031 W) x (8.64 x 104 s) ~ 4.16448 x 1036 J Therefore, Betelgeuse radiates aproximsatly 4.16 x 1036 joules in 3 significant figures of energy in one day.How much mass would need to be fully converted to energy to provide Betelgeuse with enough energy to shine for 1.00 day? Give your answer using scientific notation, the SI unit for mass and an appropriate number of significant figures. You should take the speed of light in a vacuum to be c = 3.0 x 108 m s-1. Using my calculation above I know that Betelgeuse radiates aproximsatly 4.16 x 1036 joules in SI unit and (in 3 sig-figs) of energy in 1.00 day. Using Einstein's mass-energy equivalence principle E = mc2 E is the energy (in joules) m is mass (in kg) c is the speed of light in a vacuum (3.0 x 108 m s-1) Re-arranging the equation to solve the mass. m = E Adding the numbers for c and calculating the value of E. where m: 1.16 x 1036 j (3.0 x 108 m s-1)2 = m s-1 / m2 = m-1+2 = m 1 1.16 x 1036 j (3.00 X 108 m)2 = 4.62 x 1019 kg mass Therefore, about 4.62 x 1019 kg of mass in SI unite and to three significant figuresd would need to be fully converted to energy to provide Betelgeuse with enough energy to shine for 1.00 day.The energy of stars is provided by nuclear fusion reactions. These reactions obey conservation laws, such as the conservation of total energy. In no more than two sentences, explain what is total energy and what it means for total energy to be conserved in a nuclear fusion reaction. Total energy is the sum of all potential and kinetic energy within a system. In a nuclear fusion reaction, the conservation of total energy means that the total energy before the reaction (including the mass-energy of the particles involved) is equal to the total energy after the reaction (including the kinetic energy of the products and any energy carried away by light or other particles). Give two reasons why this nuclear fusion reaction is not allowed: 18Ne JF + e~ + v, This nuclear fusion reaction is not allowed because: In nuclear reactions, the total charge must be conserved before and after the reaction. In the reaction above, the total charge on the left side is: +10 (from the Neon isotope). while on the right side, it is: +9 (from the Fluorine isotope) + -1 (from the electron, e') = +8. This violates the law of charge conservation. The lepton number is also conserved in nuclear reactions. There are no leptons on the left side of the reaction, so the lepton number is 0. On the right side, there is an electron and an electron neutrino, each with a lepton number of +1, giving a total lepton number of +2. This violates the law of lepton number conservation. Therefore, due to the violation of both charge conservation and lepton number conservation, this nuclear fusion reaction is not allowed. Complete and balance the following nuclear reactions that occur in some stars, giving the correct details for the missing entities (each indicated by ?') in the equations below: 20> BN+?2+v, & YUN+7 5130+ y 1) For this reaction 30 3N+7? +v,: 120 (oxygen-15) undergoes a nuclear reaction. It then transforms into 5N (nitrogen-15). The missing entity is a proton (p) and a neutrino(v,) is emitted. The balanced equation is: 130 N + p + v, 2) And for this reaction 2N+ ? - 120 + y 14N (nitrogen-14) undergoes a nuclear reaction. It then transforms into 130 (oxygen-15). The missing entity is a proton (p). and a photon (y) is emitted. The balanced equationis: 22N +p 150 + In this reaction, nitrogen-14 captures a proton, resulting in the formation of oxygen-15 and the emission of a gamma-ray photon (y)