A proton is trapped in an infinite square well of width 15.0 nm. The "walls" of the well are at x = 0 and x = 15.0 nm. Assuming that the system is...

Wednesday, January 28, 2015

A proton is trapped in an infinite square well of width 15.0 nm. The "walls" of the well are at x = 0 and x = 15.0 nm. Assuming that the system is...

Hello!

Yes, this probability is an integral from $\displaystyle{12}-{0.2}={11.8}{n}{m}$ to 1 $\displaystyle{2}+{0.2}={12.2}{n}{m}.$ There are two problems: what is the function to integrate and what is the value of the integral.

It is known that the function to integrate is the probability density, and it is the square of the wave function. Also it is known that the wave function for a particle in an infinite one-dimensional well with the walls $\displaystyle{x}={0}$ and $\displaystyle{x}={L}$ is

$\displaystyle\psi{\left({x}\right)}=\sqrt{{\frac{{2}}{{L}}}}{\sin{{\left(\frac{{{n}\pi{x}}}{{L}}\right)}}},$

where $\displaystyle{n}$ is the state. So we need to integrate

$\displaystyle{p}_{{d}}{\left({x}\right)}=\frac{{2}}{{L}}{{\sin}}^{{2}}{\left(\frac{{{n}\pi{x}}}{{L}}\right)}=\frac{{1}}{{L}}{\left({1}-{\cos{{\left(\frac{{{2}{n}\pi{x}}}{{L}}\right)}}}\right)}.$

It is simple, and the probability is

$\displaystyle{\int_{{{11.8}}}^{{{12.2}}}}\frac{{1}}{{L}}{\left({1}-{\cos{{\left(\frac{{{2}{n}\pi{x}}}{{L}}\right)}}}\right)}{\left.{d}{x}\right.}={\left(\frac{{x}}{{L}}-\frac{{1}}{{{2}{n}\pi}}{\sin{{\left(\frac{{{2}{n}\pi{x}}}{{L}}\right)}}}\right)}{{\mid}_{{{x}={11.8}}}^{{{12.2}}}}$

Recall that $\displaystyle{n}={2}$ and $\displaystyle{L}={15}{n}{m}$ and obtain

$\displaystyle{p}=\frac{{0.4}}{{15}}-\frac{{1}}{{{4}\pi}}{\left({\sin{{\left({4}\pi\cdot\frac{{12.2}}{{15}}\right)}}}-{\sin{{\left({4}\pi\cdot\frac{{11.8}}{{15}}\right)}}}\right)}\approx{0.048}.$

(that said, your 0.027 is the first term, 0.4/15)

Ninenotes

Wednesday, January 28, 2015

A proton is trapped in an infinite square well of width 15.0 nm. The "walls" of the well are at x = 0 and x = 15.0 nm. Assuming that the system is...

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