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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The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.
Now from Newton`s first law, the net force on gymnast,
Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (
acceleration due to gravity)
Therefore,
OR 
Given 
and
Substituting these values in above formula and calculate the force exerted by the gymnast,
Answer:
D. the rate at which work is accomplished
Explanation:
Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180
In Linear motion the swimmer swims
