Calculating Probability for a C O2 Molecule in a Closed Room

[v_pino]

Homework Statement

A C O2 molecule is released at the center of a closed room where the air is perfectly still. Take the center as the origin of coordinates. After time t has elapsed, the position of the molecule r is uncertain, but is described by the probability distribution function

f(r) = ( 1/ (4pi Dt)^3/2 ) * exp( -r^2 / 4Dt)

The diffusion coefficient of C O2 in air at 300◦ K is given by D = 1.4 × 10^−5 m2 /s. Calculate the probability to find the molecule within 0.5m of the origin after one hour.

Hint: Use spherical coordinates and convert the integral by integration by parts to one that you can calculate using the Gaussian distribution integral [say, in terms of the error function erf(x)]. Of course, you may need to use a calculator.

Homework Equations

Spherical coordinate: dV = r^2 sin (phi) d(phi) d(theta) dr

The Attempt at a Solution

Is it asking me to find <r^2>? If so, do I find it simply by integrating dr r^2 f(r) from infinity to minus infinity?

Why do I need to use spherical coordinates?

 

[vela]

No, the problem isn't asking you to find <r²>. It's asking you to find a probability.

The probability of finding the particle in an infinitesimal volume dv at the point x is given by f(x) dv, and the probability to find the particle in a volume V would then be
$$\int_V f(\vec{x})\,d^3\vec{x}$$You need to express this integral in spherical coordinates and then integrate over the proper limits (not -∞ to +∞).

 

[v_pino]

Thanks for the reply.

Here's my integral using dV in spherical coordinate system (where R = 0.5m) attached as an image. The integral of d(theta) gives me pi. But doesn't the integral of sin(phi) d(phi) give me zero?

 

[vela]

Don't see the new image, but I can make out some of the first one. I'm not sure why you have <r²> in there. That has nothing to do with the problem. You're integrating incorrectly if you get 0.

Calculate the probability to find the molecule within 0.5m of the origin after one hour.

 

[v_pino]

If the integral of sin(phi)d(phi) is zero, doesn't this mean the entire integral becomes zero? And hence, is probability zero?

Thanks

 

[vela]

Yes, and what I'm saying is that you're not evaluating the integral correctly if you're getting an answer of 0.

 

[vela]

Doesn't seem that way. In the exponential, how'd you get the 3/2 power and why isn't r squared?