Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the wave function $ \psi_n(x) $ for a particle in a rigid box is given by
$$
\psi_n(x) = \sqrt{\dfrac{2}{L}}\cdot \sin\left( \frac{n \pi x}{L} \right)\tag1
$$
where $ n = 1, 2, 3 $ represents the quantum number, $ L $ is the length of the box, and $ A $ is the normalization constant.
- For $ n = 1 $, the wave function is a single sine wave with one peak.
- For $ n = 2 $, the wave function has two peaks, showing two oscillations.
- For $ n = 3 $, the wave function has three peaks.
See the graphs below.
$$\color{blue}{\bf [b]}$$
For each value of $ n $, the particle is most likely to be found at the positions where $ |\psi_n(x)|^2 $ has its maximum values, which corresponds to the peaks of the sine wave.
- For $ n = 1 $, the peak is at $ x = L/2 $ (the center of the box).
- For $ n = 2 $, the peaks are at $ x = L/4 $ and $ x = 3L/4 $.
- For $ n = 3 $, the peaks are at $ x = L/6 $, $ x = L/2 $, and $ x = 5L/6 $.
$$\color{blue}{\bf [c]}$$
The particle is least likely to be found at the nodes of the wave function, where $ |\psi_n(x)|^2 = 0 $.
- For all $ n $, the particle is least likely to be found at $ x = 0 $ and $ x = L $ (the boundaries of the box).
- For $ n = 2 $, there's an additional node at $ x = L/2 $.
- For $ n = 3 $, there are nodes at $ x = L/3 $ and $ x = 2L/3 $, in addition to $ x = 0 $ and $ x = L $.
$$\color{blue}{\bf [d]}$$
We can analyze the probability density graphs for $ |\psi_n(x)|^2 $ to estimate the probability of finding the particle in the left one-third of the box (from $ x = 0 $ to $ x = L/3 $).
- For $ n = 1 $, the wave function is symmetric, and the probability of finding the particle in the left one-third of the box is less than $ 1/3 $ because the peak is centered at $ L/2 $, with more probability distributed around the middle.
- For $ n = 2 $, there is more probability (more than $1/3$) in the left one-third because one of the two peaks lies in this region.
- For $ n = 3 $, since one of the peaks just lies to the left one-third region, the probability will be $ 1/3 $.
$$\color{blue}{\bf [e]}$$
The probability of finding the particle in the left one-third of the box is given by the integral of $ |\psi_n(x)|^2 $ over the range $ 0 \leq x \leq L/3 $
$$
\text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) = \int_0^{L/3} |\psi_n(x)|^2 dx$$
Plug $\psi_n(x)$ from (1);
$$
\text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) = \frac{2}{L} \int_0^{L/3} \sin^2 \left(\frac{n \pi x}{L}\right) dx\tag 1
$$
To simplify the integral, we change the variable by letting:
$$u = \dfrac{n \pi x}{L} $$
This substitution gives:
$$
dx = \frac{L}{n \pi} du
$$
The new limits of integration are from $ u = 0 $ at $ x = 0 $ to $ u = \dfrac{n \pi}{3} $ at $ x = \dfrac{1}{3} L $.
Thus,
$$
\text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) = \frac{2}{n \pi} \int_0^{n \pi / 3} \sin^2(u) du
$$
$$
\int \sin^2(u) du = \frac{1}{2} \left[ u - \frac{1}{4} \sin(2u) \right]_0^ \dfrac{n \pi}{3}
$$
$$
\text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) = \frac{2}{n \pi} \left[ \frac{1}{2} \left( \frac{n \pi}{3} - \frac{1}{4} \sin \left( \frac{2n \pi}{3} \right) \right) \right]
$$
$$
\text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) = \frac{1}{3} - \frac{1}{2 n \pi} \sin \left( \frac{2n \pi}{3} \right)
$$
Therefore:
For $ n = 1 $,
$$ \text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) =\color{red}{\bf 0.195 }$$
For $ n = 2 $,
$$ \text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) =\color{red}{\bf 0.402 }$$
For $ n = 3 $,
$$ \text{Prob}\left(0 \leq x \leq \frac{1}{3} L\right) =\color{red}{\bf 0.333 }$$
