Answer:
[tex]P(0,a/3)=\frac{1}{3}-\frac{\sqrt{3}}{4\pi}[/tex]
Step-by-step explanation:
The probability of finding a particle between two points is given by:
[tex]P(x_1,x_2)=\int_{x_1}^{x_2}|\Phi_n|^2dx[/tex]
Furthermore, the wave function of a nth state is given by:
[tex]\Phi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})[/tex]
By replacing we obtain
[tex]P(0,\frac{x}{3})=\int_{0}^{\frac{a}{3}}\frac{2}{a}sin^2(\frac{n\pi x}{a})dx[/tex]
and by using the relation,
[tex]sin^2\theta=\frac{1}{2}(1-cos(2\theta))[/tex]
we get for the ground state (n=1):
[tex]P(0,a/3)=(\frac{2}{a})(\frac{1}{2})\int_{0}^{\frac{a}{3}}(1-cos(\frac{2\pi x}{a}))dx\\\\P(0,a/3)=(\frac{2}{a})(\frac{1}{2})[\frac{a}{3}-(\frac{a}{2\pi})sin(\frac{2\pi (a/3)}{a})]\\\\P(0,a/3)=(\frac{1}{a})[\frac{a}{3}-(\frac{a}{2\pi})(\frac{\sqrt{3}}{2})]=\frac{1}{3}-\frac{\sqrt{3}}{4\pi}[/tex]
hope this helps!!
