Question

An alpha particle (m = 4.002602 u) is moving at 3.60 times 10^2 m/s. a) Find its mass in kilograms. (6.64 times 10^-27 kg) b) Find its kinetic energy. (4.30 times 10^-22 J) c) Find its kinetic energy in MeV. (2.69 times 10^-9 MeV) d) Find its momentum. (2.39 times 10^-24 kg. m/s) e) Find its deBroglie wavelength (2.77 times 10^-10 m = 0.277 nm) A stationary nucleus of protactinium-231 (m = 231.035879 u) decays by alpha radiation to a nucleus with a mass of 227.027747 u. Write the full nuclear reaction. (^238_91 Pa rightarrow^4_2 He +^227_19 Ac) b) Find the energy released by this reaction. (5.151 MeV) c) If 0.679 MeV is released as heat, and assuming the actinium nucleus remains stationary, how much kinetic energy does the alpha particle have, in joules? (7.155 times 10^-13 J) d) How fast is the alpha particle moving? (1.468 times 10^7 m/s) Refer back to problem 13. a) If the alpha particle is measured as only having a speed of 7.19 times 10^6 m/s, how fast is the actinium nucleus moving? (1.27 times 10^5 m/s) b) Now how much kinetic energy does the alpha particle have? (1.07 MeV) c) How much kinetic energy does the actinium nucleus have? (0.0190 MeV) d) How much energy would be left over to be released as heat? (4.06 MeV) In a neutron capture reaction, a fast-moving neutron is absorbed into a nucleus, which then often splits into other nuclei. Aluminum-27 (m = 26.981539 u) accepts a neutron moving at 2.55 times 10^7 m/s, but then it splits into sodium-24 (m = 23.990963 u) and helium-4 (m = 4.002603 u). a) Write the nuclear reaction. (^27_13 Al +^1_0 n rightarrow^24_n Na +^4_2 He) b) How much total mass do the reactants have? (27.990204 u) c) How much total mass do the products have? (27.993566 u) d) How much kinetic energy does the neutron have, in MeV? (3.404 or 3.402 MeV) e) How much extra mass do the products have? (0.003362 u) f) How much extra energy does this represent? (3.132 MeV) g) Accounting for the energy of the incoming neutron, how much kinetic energy is left over for the reactants? (0.272 or 0.270 MeV)

Answer 1

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