Width of Finite Potential Well

[JX21]

For an electron in a certain rectangular well with a depth of 20.0 eV, the lowest
energy level lies 3.00 eV above the bottom of the well. Find the width of this well. Hint: Use tan theta = sin theta/cos theta.



Equation for energy levels in a rectangular well: (2E-V0)sin[(2mE)^(1/2)L/hbar] = 2(V0E-E^2)^(1/2)cos[(2mE)^(1/2)L/hbar]


I have attempted this problem multiple ways, and I have done similar problems giving me correct results. The simplified equation I typically use for this type of problem is derrived from the energy levels in a rectangular well equation. It is:

tan((L*(2mE)^(1/2))/(2*hbar)) = ((V0-E)/E)^(1/2)

where I can solve for L with everything else known. I converted the energies to J using 1 eV = 1.6E-19 J, the mass of electron = 9.11E-31 kg, and hbar = 1.05E-34 Js. The right hand of the equation = 51^(1/2)/3, and then inverse tan of that is 1.173. I then solve the problem and get L = 2.634E-10 m or 0.263 nm. The answer in the back of the book is 8.36 nm.
I have also tried using the Ti-89 solve function and Excel, but nothing gives me an answer even on the same order of magnitude. Can anyone see what I am doing wrong? Any help would be greatly appreciated.

 

[BigJDubs]

The correct answer is 8.36 nm. The equation for energy levels in a rectangular well is: (2E-V0)sin[(2mE)^(1/2)L/hbar] = 2(V0E-E^2)^(1/2)cos[(2mE)^(1/2)L/hbar] In order to solve for L, you need to rearrange the equation into the form of tan((L*(2mE)^(1/2))/(2*hbar)) = ((V0-E)/E)^(1/2). For this problem, you are given that the depth of the well is 20.0 eV and the lowest energy level lies 3.00 eV above the bottom of the well. This means V0 = 20.0 eV and E = 17.0 eV. Plugging these values into the equation gives you a right hand side of ((20.0 - 17.0)/17.0)^(1/2) = 0.45. Now you can use the inverse tangent function to solve for L. Inverse tangent of 0.45 is 26.18 degrees. This means L = 2 * hbar * (2mE)^(-1/2) * tan (26.18) = 8.36 x 10^(-9) m or 8.36 nm.