When an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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The energy of the electron drops when it transitions levels, as well as the atom releases photons. The emission of the photon occurs as the electron transitions from an energy state to a lower state. The photon energy represents precisely the energy that would be lost when an electron moves to a level with less energy.
When such an excited electron transitions from one energy level to another, this could emit a photon. The energy drop would be equivalent to the power of the photon that is released. In electron volts, the energy of an electron, as well as its associated photon (emitted or absorbed) has been stated.
Therefore, when an electron in a quantum system drops from a higher energy level to a lower one, the system<u> emit a photon.</u>
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Answer:
The mass defect of a deuterium nucleus is 0.001848 amu.
Explanation:
The deuterium is:
The mass defect can be calculated by using the following equation:
![\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5BZm_%7Bp%7D%20%2B%20%28A%20-%20Z%29m_%7Bn%7D%5D%20-%20m_%7Ba%7D)
Where:
Z: is the number of protons = 1
A: is the mass number = 2
: is the proton's mass = 1.00728 amu
: is the neutron's mass = 1.00867 amu
: is the mass of deuterium = 2.01410178 amu
Then, the mass defect is:
![\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20%5B1.00728%20amu%20%2B%20%282-%201%291.00867%20amu%5D%20-%202.01410178%20amu%20%3D%200.001848%20amu)
Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.
I hope it helps you!
Where is the picture to this?
1 mole = 6.22 x 10^23 molecules (Avogadro's number)
15 moles x (6.22 x 10^23) = 9.33 x 10^24 atoms
Answer:
I don't have the number of cubes in each bag, but whichever bag had the most cubes would have the most kinetic energy as it falls
