<p> Physics 1A Week 10 Questions Fall 1998</p><p>1. A coin tossed 500 times. What is the probability that the number of the heads will lie between 232 and 268? Over what range of values for the possible number of heads is there an 80% probability that the actual number of heads will fall within that range?</p><p>2. How far does a proton diffuse in water in 1 ms?</p><p>3. Molecules in a substance are allowed to diffuse over a fixed period of time. If the average collision distance could be quadrupled and the collision time held fixed, what would happen to the diffusion distance? If instead the time between collisions could be quadrupled while the velocity of the molecules is reduced to one-half, what would happen to the diffusion distance?</p><p>4. If 80% of a group of molecules lies within 1mm of their starting point after 1 ms, how long will it take for 68% of the particles to lie within 2mm of their starting point? Physics 1A Week 10 Solutions Fall 1998</p><p>1. If the number of heads lies between 232 and 268, then the difference between the number of heads and the number of tails is less than or equal to 36. Now 36 = 1.61 sqrt(500), which is close to 1.65 sqrt(500), so the probability is close to 90% (actually a little bit less). For the second question, we would need to have the difference between the heads and the tails less than 1.28 sqrt(500), which is 29. The desired range is therefore approximately between 236 and 264.</p><p>2. The formula is x(rms) = sqrt(2Dt). The value of D comes from the table on p. 31, which gives D = 9 X 10^-9 m^2/s for hydrogen ions in water. Substituting in, we get about 4.2 m.</p><p>3. The relevant formula is x(rms) = sqrt(^2 * t/), so the distance is proportional to , and it quadruples. For the second question, we have  = v , so if  quadruples and v is halved, then  doubles. Then x(rms) will remain un- changed.</p><p>4. Distance diffused = 1.28 sqrt(N1) , 1.28 sqrt(N1)  = 1mm, where N1 is the number of collisions for this case. We also want sqrt(N2)  = 2mm. So (sqrt(N2) ) /(1.28 sqrt(N1) ) = 2mm / 1mm = 2, which implies sqrt(N2/N1) = 2.56, or N2 = 6.55 N1. So the number of collisions increases by a factor of 6.55, and the total time must increase by 6.55, since collision time is fixed. So t=6.55 ms.</p>