# 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 this is correct but is this simply 1/`sqrt(L/2)`...

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 this is correct but is this simply 1/`sqrt(L/2)`

=.559 mm^-1/2` `

b) What is the probability of finding the electron in a 0.010 nm-wide region centered at x = 1.0 nm? Is this simply .00228630*100 = .23%

c) What is the probability of finding the electron in the interval 1.15 nm `<= ` x `<=` 1.84 nm. I don't know if this is correct but is this simply .0025189*100 = .25%``

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Hello!

`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 `int_c^d pd(x) dx.` In our case for positive c and d it is

`b^2 int_c^d e^(-(2x)/L) dx = b^2*L/2*(e^(-(2c)/L) - e^(-(2d)/L)).`

a) the value of b must be such that the total probability, `int_(-oo)^(+oo) pd(x) dx,` = 1. In our case it is `int_0^(+oo) b^2 e^(-(2x)/L) dx = b^2*L/2 = 1.`

So yes, `b=sqrt(2/L)` and for L=6.4 it is about `0.559 ((nm)^(-1/2)).`

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")