Question

Using the half-lives for 238U and 235U, the natural abundance of each isotope, and knowing the density of uranium metal is 19.2 g/cm3, calculate the total decay rate (disintegrations per second) of uranium atoms in a solid sphere, 10 cm in diameter. Both 238U and 235U decay by alpha emission. The alpha particle energies are approximately 4.3 MeV and 4.7 MeV, respectfully. There is additional energy released in photons (gammas) and subsequent decay of daughter products. Ignoring all that, using only the primary alpha particle energies, calculate the heat generated in the sphere of uranium in watts.

Answer 1

Density of uranium metal is 19.2 g/em^3 therefore Mass of a solid sphere of diameter 10 em on radius 5 em of uranium metal m = (4/3 times pi times 5^3 times 19.2)g Now, Natural uranium exists of 99.3% U^238 and 0.70% U^235 therefore Mass of U^238 in the sphere m_238 = (m times 99.3/100)g Atomic mass of U^238 = 23.8gm therefore In 238 gm of U^238 no. of atom = 6.023 times 10^23 therefore In (m times 99.3/100)gm of U^238 no. of atom N_238 = (6.023 times 10^23/238 times m times 99.3/100) mass of U^235 in the sphere, m_235 = (m times 0.70/100)g \r\n therefore No. of atom present in m_235 gm of U^235 = (6.023 times 10^23/235 times m times .70/100) = N_235 Now, Decay Rate is defined by, R = lambda N where, lambda = disintegration constant N = no. of atom of sample present. If t_1/2 is the half