Part (a) of the figure below shows the energy-level diagram for a finite, one-di
ID: 1973594 • Letter: P
Question
Part (a) of the figure below shows the energy-level diagram for a finite, one-dimensional energy well that contains an electron. The nonquantized region begins at E4 = 550.0 eV. Part (b) of the figure below gives the absorption spectrum of the electron when it is in the ground state - it can absorb at the indicated wavelengths: a = 13.220 nm and b = 4.7674 nm and for any wavelength less than c = 2.8914 nm. What is the energy of the first excited state? (in eV)
Please show your work.
Please calculate a final answer.
Explanation / Answer
Reading between the lines (no pun intended) we can see thatthe energy associated with ?c must beequal to the difference between E0 and E4. This is becauseany photon of wavelength less than ?c is enoughto boost the electron from E0 (ground state) into the"nonquantized" region.
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So... find the energy associated with?c:
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E = h c / = 4.14 x10^-15 * 3 x 10^8 / 2.8914 x10^-9 = 429.5497 eV
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Then... E4 - E0 = 429.5497
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550.0 - E0 = 429.5497
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E0 = 120.4503 eV (this is ground state energy)
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Now the energy of the first excited state is the ground stateenergy plus the energy from the least energetic (longestwavelength) photon that can be absorbed:
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E = hc / = 4.14 x 10^-15 * 3 x 10^8 /13.220 x 10^-9 = 93.9486 eV
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So the energy of the first excited stateis 93.9486 + 120.4503 = 214.3989 eV