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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Answer:the pressure depends on gas and it will be half as much underwater
Explanation:
Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.
This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .
Kinetic energy is the energy for a catapult.
Answer:
For C1, Q = 1.6125×10⁻³ C
For C2, Q = 6.25×10⁻⁴ C
Explanation:
Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.
From the question,
Q = CV........................ Equation 1
Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.
For the first capacitor,
Q = C1V............. Equation 2
Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V
Substitute into equation 2
Q = (6.45×10⁻⁶ )(250)
Q = 1.6125×10⁻³ C.
For the the second capacitor,
Q = C2V............. Equation 3
Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V
Q = (2.5×10⁻⁶ )(250)
Q = 6.25×10⁻⁴ C
