"

Set 57 Problem number 16


Problem

A hypothetical atom with negligible kinetic energy has a mass of 230.889 amu. It undergoes an alpha decay. The remaining atom has atomic mass 226.8831 amu. What is the kinetic energy and/or wavelength (whichever is more appropriate) of the emitted particle, assuming that the kinetic energy of the remaining atom is negligible? How much energy would be released by the decay of a mole of these atoms? Note that the mass of a helium atom is about 4.0026 amu, where an amu is approximately 1.66 * 10^-27 kg.

Solution

The change in atomic mass is approximately 230.889 amu - ( 226.8831 amu + 4.0026 amu) = 3.350735E-03 amu, or about 5.56222E-03 * 10^-27 kg.

This corresponds to an energy of E = m c^2 = 5.56222E-03 * 10^-27 kg ( 3 * 10^8 m/s) ^ 2 = 3.015662 * 10^-13 Joules.

A mole of these nuclei would constitute 6.02 * 10^23 nuclei, each releasing 3.015662 * 10^-13 Joules. The total energy released would therefore be

"