Question

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

Answer 1

The kinetic energies and speeds of the two helium atoms are 92.2 keV and $1.49*10^6 m/s$

Further explanation

An atom a fundamental piece of matter. It is made up of three tiny kinds of particles called subatomic particles such as protons, neutrons, and electron.

Beryllium is a steel-gray metal that is brittle at room temperature.

The kinetic energy of an object is the energy that possesses due to its motion.

A helium atom is an atom of the chemical element helium. Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with one or two neutrons depending on the isotope and held together by the strong force.

Let the two helium atom move in opposite direction along the x axis speed $V_1$ and $V_2$. Convention of momentum along with the $x$ direction. ($P_x, initial = P_x, final$) given

$0=m_1V_1-m_2V_2 on V_1=V_2$

The energy is in the form of the total kinetic energy of the two helium atoms

$H_1+H_2=92.2 keV$

Because $V_1=V_2$, it follows that $H_1=H_2=46.1 keV$ therefore

$v = \frac{2*H_1}{m_1} = \sqrt{\frac{2*(46.1*10^3eV)(1.602*10^{-19} J/eV)}{(4.004)*(1.6605*10^{-27}kg/m)} } = 1.49*10^6 m/s$

$V_2=V_1=1.49*10^6 m/s$

Learn more

1. Learn more about beryllium brainly.com/question/1554284
2. Learn more about helium brainly.com/question/2882934
3. Learn more   about energy brainly.com/question/9736685

Answer details

Grade:    9

Subject:  chemistry

Chapter:  atom

Keywords: beryllium, helium, energy, the kinetic energies, speeds

Answer 2

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