Answer:
ΔU = 2 mg h
Explanation:
In a spring mass system the potential energy is U = m g h
where h is measured from the equilibrium point of the spring
the potential energy at the highest point is
U₁ = m g h
the potential energy at the lowest point is
U₂ = m g (-h)
instead in this energy it is
ΔU = 2 mg h
In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero
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:
M=28.88 gm/mol
Explanation:
Given that
T= 95 K
P= 1.6 atm
V= 4.87 L
m = 28.6 g
R=0.08206L atm .mol .K
We know that gas equation for ideal gas
P V = n R T
P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant
Now by putting the values
P V = n R T
1.6 x 4.87 = n x 0.08206 x 95
n=0.99 moles
We know that number of moles given as
M=Molar mass
M=28.88 gm/mol
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