In summary, The two parts (a) and (b) are not the same, as they are asking for different values of the particle's position. Part (a) is asking for the average position, which can be calculated by finding the area under the probability distribution curve. Part (b) is asking for the most likely position, which can be found by locating the maximum of the probability distribution curve. In order to solve these problems, integration by parts can be used to evaluate the integrals.
[itex]x \to \infty[/itex] is one extremum, but certainly not a maximum...[itex]P(x \to \infty)=0[/itex]...There is one more solution to that equation which does correspond to a maximum.
Don't forget to express [itex]\langle x \rangle[/itex] in terms of [itex]\beta[/itex] in order to properly compare the average value to the expected value
the reason for this is that the wave function has a long 'tail' which makes more 'probability' lie to the right of its maximum (c.f Maxwell speed distribution)
#85
TFM
1,026
0
Indeed, not much difference, just *half more
Would the Standard deviation for this be the difference, or will I need to use:
so now you added a question to the original one, how rude!
and I/we don't have to feed you like a baby, have some confident
[tex]
\int x^2e^{-2\beta x} \, dx
[/tex]
now you know what the integral is, you did earlier