A beryllium-8 nucleus at rest undergoes double alpha decay as follows:
$^{8}_{4}$ Be → $^{4}_{2}$ He + $^{4}_{2}$ He The known atomic masses are:
$^{4}_{2}$ He: 4.002603 $^{8}_{4}$ nd{array}\) Be: 8.005305 u
What is the kinetic energy of each departing alpha particle?
(1 u = 931.5 MeV/c²)

Question

A beryllium-8 nucleus at rest undergoes double alpha decay as follows:

^{8}_{4}

Be →

^{4}_{2}

He +

^{4}_{2}

He The known atomic masses are:

^{4}_{2}

He: 4.002603

^{8}_{4}

nd{array}\) Be: 8.005305 u
What is the kinetic energy of each departing alpha particle?
(1 u = 931.5 MeV/c²)

Quizplus · Accepted Answer

The correct answer is A.The total kinetic energy released in the decay is:$$Q = (m_{Be} - 2m_{He})c^2$$$$Q = (8.005305 u - 2(4.002603 u))c^2$$$$Q = 0.004299 u c^2$$$$Q = (0.004299 u)(931.5 MeV/c^2)$$$$Q = 4.00 MeV$$Each alpha particle receives half of this energy, so the kinetic energy of each departing alpha particle is:$$K = Q/2 = 2.00 MeV$$$$K = (2.00 MeV)(1.602 × 10^{−13} J/MeV)$$$$K = 3.20 × 10^{−13} J$$$$K = 3.20 × 10^{−13} J(1 keV/1.602 × 10^{−16} J)$$$$K = 200 keV$$Therefore, the kinetic energy of each departing alpha particle is 200 keV, which is closest to choice A, 46 keV.

A Beryllium-8 Nucleus at Rest Undergoes Double Alpha Decay as Follows