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Homework Statement
An atom that is being monitored emits light five times without being re-excited. The energies detected for four of those emissions are 0.7 eV, 0.8 eV, 0.9 eV, and 2.0 eV. The energy information for the other emission was lost by the computer controlling the detectors, as was the information about the sequence of detection. The lowest 12 energy levels of the atom, in electron volts, are
6.5 4.1 2.6
5.3 3.8 2.0
4.9 3.4 1.5
4.5 2.9 0
(a) From which energy level did the atom begin its quantum jumping downward in energy? (b) What value of energy was lost by the computer?
Homework Equations
N/A
The Attempt at a Solution
The total energy of the four known emissions is 4.4 eV.
4.1 eV and under are all less than 4.4 eV so the starting point must be at the 4.5 eV level or higher.
If the starting point were at 4.5 eV then the missing jump would be 0.1 eV. Since there is no jump of 0.1 eV anywhere in the sequence, this cannot be the starting point.
4.9 eV works as a starting point. The missing energy value would be 0.5 eV and the jumps would be 4.9 eV to 4.1 eV, (0.8 eV), 4.1 eV to 3.4 eV, (0.7 eV), 3.4 eV to 2.9 eV,(0.5 eV), 2.9 eV to 2.0 eV,(0.9 eV) and finally, 2.0 eV to 0 eV.
5.3 eV also works with reasoning similar to the above solution with a missing jump of 0.9 eV.
The answer in the back of the book is 5.3 eV, with a missing jump of 0.9 eV.
Is this a case of two correct answers and only one listed, or am I making an error for the 4.9 eV starting point?
Thanks for any help.