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Here is given, width of well = a

Position (x) = \( \frac {a}{4} \)

Normalized wave function \( \Psi \) (x,t) = \( \sqrt \frac {2}{a} \)sin\( \frac {n\pi x}{a} \) \(e^{- \frac {iEt}{ℏ}}\)

We know that, probability of finding a particle, P = \( \Psi\times\Psi \)

Then, at x = \( \frac {a}{4} \)

P = \( \left ( \sqrt \frac {2}{a} sin\frac{n\pi}{a}.\frac{a}{4}e^{\frac{iEt}{ℏ}} \right )\)\( \left ( \sqrt \frac{2}{a}sin\frac{n\pi}{a}.\frac{a}{4}e^{-\frac{iEt}{ℏ}} \right )\)

P = \( \frac{2}{a}sin^{2}\frac{n \pi}{4} \)…………………. (1)

If n = 1, P_{1} = \( \frac{2}{a}sin^{2}\frac{\pi}{4} = \frac {1}{a}\)

For n = 2, P_{2} = \(\frac{2}{a} sin^{2}\frac{2\pi}{4} = \frac{2}{a}\)

And for n = 3, P_{3} = \(\frac{2}{a}sin^{2}\frac{32\pi}{4} = \frac{2}{a}sin^{2}135^{\circ} = \frac{2}{a}\left ( \frac{1}{\sqrt2} \right )^{2} \)

∴ P_{3} = \( \frac{1}{a}\)

Hence, the probability of finding a particle in a well of width of a position x = \(\frac{a}{4} \) form the wall for n =1, n = 2 and n = 3 are \(\frac{1}{a},\frac{2}{a} \enspace and \enspace \frac{1}{a} \) respectively.

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