Hi there!
Use the equation:
Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:
v2 = ?
m1 = 0.048 kg (converted)
m2 = 2.95
v1 = 391
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
Drop "moves" from the list for a moment.
You can also drop "stops moving", because that's included in "changes speed"
(from something to zero).
When an object changes speed or changes direction, that's called "acceleration".
I dropped the first one from the list, because an object can be moving,
and as long as it's speed is constant and it's moving in a straight line,
there's no acceleration.
I think you meant to say "starts moving". That's a change of speed (from zero
to something), so it's also acceleration.
