An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the original beryllium atom is at rest, find the kinetic energies and speeds of the two helium atoms

The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s

The density would be 1.622