Question

The probability that an atom moving with speed v sutters a collision with another atom in the very next interval of time dt is vdt/e, where is a constant. (a) Show that the probability that an atom travels a distance svt in time t without undergoing any collisions is e (b) The probability that the atom undergoes its first collision in the distance interval between s and s + ds is P(s)ds. Find P(s) (c) Find 〈s) and 〈say in terns of 1. (d) Find the mean squared distance from the starting point after N collisions. Assume each collision randomizes the direction of motion of the atom. Useful integral: 0

Answer 1

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   Given that, an atom moving with speed v suffers a collision with another atom in the very next interval of time dt is given by, Vdt/l a) In this time dt the distance travelled by the atom = vdt = ds so, the probability of collision reduces to ds/l Now the probability that the atom travels the distance ds without suffering collision is = (1 - ds/l) [since seem of total probability = 1] now, the probability that the atom goes a distance s without suffering collision = f(s) so, a molecule will travel a further distance (s + ds) without any collision is given by f(s + ds) = f(s) (1 - ds/l) equation (1) using Taylor expansion for for f(s + ds) f(s + ds) = f(s) + f^'(s) middot ds(ignoring higher orders since they are negligible \r\n so (1) reduces to so (1) reduces to f(s) + f^'(s) ds = f(s) - f(s) ds/L f^'(s) ds = -f(s)
   
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