The wave function for an electron that is confined to x `>=` 0 nm See the image below. a) What must be the value of b? I don't know if...

Tuesday, June 21, 2011

The wave function for an electron that is confined to x $\displaystyle\ge$ 0 nm See the image below. a) What must be the value of b? I don't know if...

Hello!

$\displaystyle\psi$ is the standard symbol for a wave function. Its square is the probability density pd(x). By the definition of probability density, the probability of being between c and d is $\displaystyle{\int_{{c}}^{{d}}}{p}{d}{\left({x}\right)}{\left.{d}{x}\right.}.$ In our case for positive c and d it is

$\displaystyle{{b}}^{{2}}{\int_{{c}}^{{d}}}{{e}}^{{-\frac{{{2}{x}}}{{L}}}}{\left.{d}{x}\right.}={{b}}^{{2}}\cdot\frac{{L}}{{2}}\cdot{\left({{e}}^{{-\frac{{{2}{c}}}{{L}}}}-{{e}}^{{-\frac{{{2}{d}}}{{L}}}}\right)}.$

a) the value of b must be such that the total probability, $\displaystyle{\int_{{-\infty}}^{{+\infty}}}{p}{d}{\left({x}\right)}{\left.{d}{x}\right.},$ = 1. In our case it is $\displaystyle{\int_{{0}}^{{+\infty}}}{{b}}^{{2}}{{e}}^{{-\frac{{{2}{x}}}{{L}}}}{\left.{d}{x}\right.}={{b}}^{{2}}\cdot\frac{{L}}{{2}}={1}.$

So yes, $\displaystyle{b}=\sqrt{{\frac{{2}}{{L}}}}$ and for L=6.4 it is about $\displaystyle{0.559}{\left({{\left({n}{m}\right)}}^{{-\frac{{1}}{{2}}}}\right)}.$

And the formula for a probability becomes

for positive c and d. If c is negative, c must be replaced with zero.

b) use this formula for c=1-0.005 and d=1+0.005.

c) use this formula for c=1.15 and d=1.84.

(there is an error at the picture, must be "for x>=0 nm", not "for x>=nm")

Ninenotes

Tuesday, June 21, 2011

The wave function for an electron that is confined to x $\displaystyle\ge$ 0 nm See the image below. a) What must be the value of b? I don't know if...

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Tuesday, June 21, 2011