A particle of mass $m$ in an infinite potential well of width $a$ is subjected to a perturbation, $V' = \frac{h^2}{40ma^2}$ as shown in figure, where $h$ is Planck's constant. 
The first order energy shift of the fourth energy eigenstate due to this perturbation is
$(\frac{h^2}{Nma^2})$
The value of $N$ is ____________ (in integer).
1. Identify the wavefunction for the $n$-th energy state:
For an infinite potential well of width $a$ defined from $x = 0$ to $x = a$, the normalized wavefunctions are given by:
$\psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right)$
For the fourth energy eigenstate ($n = 4$):
$\psi_4(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{4\pi x}{a}\right)$
2. Set up the first-order energy shift formula:
According to first-order perturbation theory, the energy shift $E_n^{(1)}$ is the expectation value of the perturbation $V'$ in the unperturbed state:
$E_n^{(1)} = \int_0^a \psi_n^*(x) V'(x) \psi_n(x) \, dx$
From the figure, $V'(x)$ is non-zero only in the region from $x = \frac{3a}{4}$ to $x = a$:
$E_4^{(1)} = \int_{3a/4}^a \left( \sqrt{\frac{2}{a}} \sin\left(\frac{4\pi x}{a}\right) \right)^2 \cdot V' \, dx$
$E_4^{(1)} = \frac{2V'}{a} \int_{3a/4}^a \sin^2\left(\frac{4\pi x}{a}\right) \, dx$
3. Evaluate the integral:
Use the trigonometric identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$:
$\int_{3a/4}^a \sin^2\left(\frac{4\pi x}{a}\right) \, dx = \int_{3a/4}^a \frac{1 - \cos(8\pi x/a)}{2} \, dx$
$= \frac{1}{2} \left[ x - \frac{a}{8\pi} \sin\left(\frac{8\pi x}{a}\right) \right]_{3a/4}^a$
$= \frac{1}{2} \left[ \left( a - 0 \right) - \left( \frac{3a}{4} - \frac{a}{8\pi} \sin(6\pi) \right) \right]$
$= \frac{1}{2} \left[ a - \frac{3a}{4} \right] = \frac{1}{2} \left( \frac{a}{4} \right) = \frac{a}{8}$
4. Calculate $E_4^{(1)}$ and find N:
Substitute the value of the integral and $V' = \frac{h^2}{40ma^2}$ back into the equation:
$E_4^{(1)} = \frac{2}{a} \cdot \left( \frac{h^2}{40ma^2} \right) \cdot \frac{a}{8}$
$E_4^{(1)} = \frac{2h^2}{320ma^2} = \frac{h^2}{160ma^2}$
Comparing this with the given form $\frac{h^2}{Nma^2}$:
$N = 160$
The value of N is 160.
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