Answer:
a.
445 nm : 4.45 x 10^-19 J
445 nm : 2.78 eV
586 nm : 3.38 x 10^-19 J
586 nm : 2.11 eV
667 nm : 2.99 x 10^-19 J
667 nm : 1.87 eV
b.
E3 to E1 : ΔE = 1.87 eV
E3 to E2 : ΔE = 0.24 eV
E2 to E1 : ΔE = 0.24 eV
Step-by-step explanation:
a) To calculate the energies of the emitted photons, we can use the formula:
E = hc/λ
where E is the energy of the photon in Joules, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon in meters.
For the first line of wavelength 445 nm:
E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(445 x 10^-9 m) = 4.45 x 10^-19 J
To convert this to electron volts (eV), we can use the conversion factor 1 eV = 1.602 x 10^-19 J:
E = 4.45 x 10^-19 J/(1.602 x 10^-19 J/eV) = 2.78 eV
For the second line of wavelength 586 nm:
E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(586 x 10^-9 m) = 3.38 x 10^-19 J
E = 3.38 x 10^-19 J/(1.602 x 10^-19 J/eV) = 2.11 eV
For the third line of wavelength 667 nm:
E = hc/λ = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(667 x 10^-9 m) = 2.99 x 10^-19 J
E = 2.99 x 10^-19 J/(1.602 x 10^-19 J/eV) = 1.87 eV
b) The energy level diagram for an atom of the element which has produced these photons can be drawn as follows:
markdown
E3
|
| ΔE = 1.87 eV
|
|
|
E2
|
| ΔE = 0.24 eV
|
|
|
E1
The electron transitions which have given rise to the three spectral lines are:
The transition from energy level E3 to E1, which produces the photon of wavelength 445 nm and energy 2.78 eV (ΔE = 1.87 eV).
The transition from energy level E3 to E2, which produces the photon of wavelength 586 nm and energy 2.11 eV (ΔE = 0.24 eV).
The transition from energy level E2 to E1, which produces the photon of wavelength 667 nm and energy 1.87 eV (ΔE = 0.24 eV).
Note that this energy level diagram is just one possible arrangement of energy levels that could produce the observed spectral lines. There may be other possible arrangements of energy levels that could also explain the observed lines.
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