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v=Pa/m (c) Determine the total energy released during the decay of 1 mole of bismuth 212: (5.98 x 10¹¹J)

v=Pa/m (c) Determine the total energy released during the decay of 1 mole of bismuth 212: (5.98 x 10¹¹J)

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how would you do part c? this is a non graded practice worksheet
1. A bismuth isotope decays into thallium by emitting an alpha particle according to the following equation: + 217 mass Atomic #83 only protons. Bi → TI + He (a) Determine the atomic number Z and the mass number A of the thallium nuclei produced and enter your answers in the spaces provided below. Z = 81 A 213 The mass of the alpha particle is 6.64 x 10-27 kg. Its measured kinetic energy is 6.09 MeV and its speed is much less than the speed of light. 46.09 ebev 9.744e-13 J (a) Determine the momentum of the alpha particle. (1.14 x 10-¹9 kg m/s) K=//mv² 9.744e-13-1 (6.64x10-27)√² P=mv v=1.713e7 m/s. articles (b) Determine the kinetic energy of the recoiling thallium nucleus. (1.93 x 10-¹4J) LSS == 213 0 = 1.66 X10-27 (213) = 353,58e-27 kg P--Pa mass V=1.1386-199s from 353.58e-27 my-la 0 PILYS dicas K = 1/2mv ² ====₂( -1.84е-145 ST (c) Determine the total energy released during the decay of 1 mole of bismuth 212: (5.98 x 10¹¹J) v=Pa/m P=6164x10-27 (1.713e7) P =1.138e-19 kgm/s/ =3.224 15m/s dont on the metal
Transcribed Image Text:1. A bismuth isotope decays into thallium by emitting an alpha particle according to the following equation: + 217 mass Atomic #83 only protons. Bi → TI + He (a) Determine the atomic number Z and the mass number A of the thallium nuclei produced and enter your answers in the spaces provided below. Z = 81 A 213 The mass of the alpha particle is 6.64 x 10-27 kg. Its measured kinetic energy is 6.09 MeV and its speed is much less than the speed of light. 46.09 ebev 9.744e-13 J (a) Determine the momentum of the alpha particle. (1.14 x 10-¹9 kg m/s) K=//mv² 9.744e-13-1 (6.64x10-27)√² P=mv v=1.713e7 m/s. articles (b) Determine the kinetic energy of the recoiling thallium nucleus. (1.93 x 10-¹4J) LSS == 213 0 = 1.66 X10-27 (213) = 353,58e-27 kg P--Pa mass V=1.1386-199s from 353.58e-27 my-la 0 PILYS dicas K = 1/2mv ² ====₂( -1.84е-145 ST (c) Determine the total energy released during the decay of 1 mole of bismuth 212: (5.98 x 10¹¹J) v=Pa/m P=6164x10-27 (1.713e7) P =1.138e-19 kgm/s/ =3.224 15m/s dont on the metal
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