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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The answer is 1,600 J.
A work (W) can be expressed as a product of a force (F) and a
distance (d):
W = F · d<span>
We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
The formula is F = ( q1 * q2 ) / r ^ 2
<span>where: q is the individual charges of each ion </span>
<span>r is the distance between the nuclei </span>
<span>The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. </span>
<span>From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. </span>
<span>For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. </span>
<span>So Far: ( BaO and SrO ) > ( NaF and KCl ) </span>
<span>The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. </span>
<span>For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. </span>
<span>Therefore: </span>
<span>Answer: SrO > BaO > NaF > KCl </span>
