The IMA of the pulley shown is 2.
The answer would be 2.63. Your welcome. This has been changed to the correct answer.
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.
Learn more about energy level here: brainly.com/question/14287666
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Answer:
Wouldn’t his or hers speed be 10m?
Explanation:
because 60 divided by 6 = 10
so 10m per second?
Answer:
2.55 Hz
Explanation:
frequency = Speed of wave / wavelength
= 21.2 / 8.3
= 2.55 Hz
