The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
<h3>Change in energy level of the electron</h3>
When photons jump from a higher energy level to a lower level, they emit or radiate energy.
The change in energy level of the electrons is calculated as follows;
ΔE = Eb - Ef
ΔE = -2.68 eV - (-5.74 eV)
ΔE = 3.06 eV
Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
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E=mc² where c is speed of the light
3 m/s more andmore less than speed of the light. So mass of the person still 100 kg
Answer:
9.22 s
Explanation:
One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad
Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:
For the child: 
For the horse: 
For the child to catch up with the horse, they must cover the same angular distance within the same time t:



t = 25.05 or t = 9.22
Since we are looking for the shortest time we will pick t = 9.22 s
Answer:
Approximately
, assuming that the volume of these two charged objects is negligible.
Explanation:
Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.
Let
and
denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question,
and
.
Let
denote the distance between these two point charges. In this question,
.
Let
denote the Coulomb constant. In standard units,
.
By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:
.
Substitute in the values and evaluate:
.
100% C . By size and distance