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Nonamiya
1 month ago
10

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 or

iginal beryllium atom is at rest, find the kinetic energies and speeds of the two helium atoms
Chemistry
2 answers:
lorasvet [2.7K]1 month ago
6 0

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 is a fundamental unit of matter. It consists of three small types of particles known as subatomic particles, including protons, neutrons, and electrons.

Beryllium, a steel-gray metal, tends to be brittle at standard room temperature.

The kinetic energy represents the energy an object has due to its motion.

A helium atom is defined as an atom of the chemical element helium. Helium comprises two electrons held together by electromagnetic forces and a nucleus of two protons, along with one or two neutrons based on the isotope, bound together by strong forces.

The two helium atoms are expected to move apart on the x-axis with speeds V_1 and V_2. This adheres to momentum conservation along the x direction. (P_x, initial = P_x, final) given

0=m_1V_1-m_2V_2 on V_1=V_2

The released energy translates into the total kinetic energy of the two helium atoms

H_1+H_2=92.2 keV

Provided that V_1=V_2, this leads to 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
  2. Learn more about helium
  3. Learn more   about energy

Answer details

Grade:    9

Subject:  chemistry

Chapter:  atom

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

VMariaS [2.9K]1 month ago
4 0
The energy released results in a kinetic energy of 92.2 keV for the products. We should convert keV into Joules, noting that 1 keV equals a kiloelectron volt. The required conversion is: 1.602×10⁻¹⁹ <span>joule = 1 eV

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

Next, we can determine the velocity of each He atom from the kinetic energy:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
This solves to give us: v = 5.367×10¹¹ m/s
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Answer:

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