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An electron is confined within a tube of length L. The potential energy of the electron is zero in the first half of the tube and 12 eV in the second half. The total energy of the electron is 16 eV. Find the ratio of the longest to the shortest de Broglie wavelength of the electron inside the tube.
- 2
- sqrt(2)
- 4
- sqrt(2)/2
Correct answer: 2
Solution
Since lambda is proportional to 1/sqrt(KE), the ratio of longest (lower KE) to shortest (higher KE) wavelength is sqrt(16)/sqrt(4) = 4/2 = 2.
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