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
#SPJ1
<h2>
The balloon is moving when it is halfway down the building at 20.78 m/s.</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 0.5 x 44 = 22 m
Substituting
v² = u² + 2as
v² = 0² + 2 x 9.81 x 22
v² = 431.64
v = 20.78 m/s
Velocity at 22 m = 20.78 m/s
The balloon is moving when it is halfway down the building at 20.78 m/s.
I believe that number 2 is A, not sure though.
