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.
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:
connect two 9 ohms resistance in series now it becomes 18 ohm
length
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The answer is C. Click C.
