Electron Energy Levels in Nanowire | Approximation & Quantum #s

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The discussion revolves around approximating an electron in a nanowire as a one-dimensional infinite square-well potential, specifically for a 2 um long wire at 13K. The calculated energy for the first quantum level (n=1) is confirmed as 9.40 x 10^-8 eV. However, confusion arises regarding the second energy level, where the participant believes it should be four times the first level's energy (n=2 leading to E2 = 4E1). The book claims it is only twice the first energy level, leading to a disagreement on the correct calculation. This discrepancy highlights the need for clarity on energy level equations in quantum mechanics.
Pengwuino
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The question is:

We can approximate an electron moving in a nanowire as a one-dimensional infinite square-well potential. Let the wire be 2 um long. The nanowire is cooled to a temperature of 13K, and we assume the electron’s average kinetic energy is that of a gas molecule at this temperature ( = 3/2 kT). (a) What are the three lowest possible energy levels of the electrons? (b) What is the approximate quantum number of the electrons moving in the wire?

So I have the equation to find the energy in an infinite well…

\[<br /> \begin{array}{l}<br /> E_n = n^2 \frac{{\pi ^2 (\hbar c)^2 }}{{2mc^2 L^2 }} \\ <br /> L = 2*10^3 nm \\ <br /> m = 5.11*10^5 eV \\ <br /> \hbar c = 197.33eV*nm \\ <br /> \end{array}<br /> \]

Using this, I find n=1 to equal 9.40*10^-8 eV

The book says I am good there.

The problem is when I have to find the second energy level. I figured you use n=2 which means its just 2^2 or 4 times the energy but the book says its just 2 times the energy of the first energy level. What am I not understanding here?
 
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I'm pretty sure the book made a mistake. For n=2, E2 = 4*E1 = 3.8 X 10^-7. I double checked with cramster, and that's how it was done there also.
 

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