# The wave function for a particle is `psi(x)=Axe^(-x^2/a^2)` where A and a are constants. Where is the particle most likely to be found? Assume that a = 2.49 nm.   Is the particle most likely around 2.49 nm because that is the constant/max.

Hello!

The probability of finding a particle within some set `R` is `int_R |Psi(x)|^2 dx.` The probability to find a particle at a specific point is zero, but there is a correct question: "what is the point `x_0` such that the probability of finding the particle within a small interval with the center in `x_0` is maximal?"

Since our `Psi(x)` is continuous, the integral over a small interval is almost equal to `Delta x*|Psi(x_0)|^2.` So, we have to find the point(s) `x_0` where `|Psi(x_0)|^2` has its maximum. This is the same as the `|Psi(x_0)|` maximum.

The factor `|A|` has no effect on `x_0,` thus it is sufficient to find the maximum of `f(x) = x e^(-x^2/a^2)` for `xgt=0` (for `xlt0` the values are the same). At `x=0` the value is zero, at `+oo` the limit is also zero, so the maximum is somewhere in between. The necessary condition is `f'(x_0) = 0,` so the equation is:

`f'(x) = (x e^(-x^2/a^2))' =e^(-x^2/a^2) - x*(2x)/a^2e^(-x^2/a^2) =e^(-x^2/a^2)(1-(2x^2)/a^2) = 0.`

The only such `x_0 = |a|/sqrt(2)` (so there are two points of a maximum, `|a|/sqrt(2)`  and `-|a|/sqrt(2)`). Numerically for `a=2.49 nm`   `x_0 approx +-1.76 nm.`

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