Probability of finding a particle?

[iafatel]

Homework Statement

For a particle in the ground state of a rigid box, calculate the probability of finding it between x=0 and x=\frac{L}{3}

Homework Equations

\left|\psi^{2}\right| = \frac{2}{L}Sin^{2}\left(\frac{nx\pi}{L}dx\right)

The Attempt at a Solution

= \frac{2}{L}\int^{\frac{L}{3}}_{0} Sin^{2}\left(\frac{x\pi}{L}\right)
= \frac{2}{L}\int^{\frac{L}{3}}_{0} \frac{1-Cos \left(\frac{2x\pi}{L}\right)}{2}

\frac{1}{L} \int^{\frac{L}{3}}_{0} 1 - \int^{0}_{\frac{L}{3}} Cos \left(\frac{2x\pi}{L}\right)}

\frac{1}{L} \left[ \left|x\right|^{\frac{L}{3}}_{0} - \left|\frac{L}{2\pi}Sin\left(\frac{2x\pi}{L}\right)\right|^{\frac{L}{3}}_{0}\frac{1}{L} \left[ \frac{L}{3}} - \frac{L}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right]

factor out the L

\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right]

stuck here...when I work it out I get a negative number...am I mising something?

 

[Mentz114]

\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right]

I get a positive number -

sin(2pi/3) = 0.8660

0.8660/(2*pi ) = 0.1378

1/3 - 0.1378 = 0.1955

 

[iafatel]

oh wow...stupid math mistake...forgot the parenthesis on my calc =(

thank you!

 

[tazgurl78]

what would the probability of the particle be if x=1.95 and 2.05?

 

[tazgurl78]

the answer is supposed to be .007, but I keep getting -.2921

 

[sergioeldiego]

I think you should be using cos instead of sin for the ground state.