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For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Sunday, December 5, 2010
Chapter 29+30, Problems 7-12
Material to be covered in the last day of class, December 9th.
Due Monday, Dec. 13th, at midnight.
17 comments:
Anonymous
said...
I am having trouble with #10. I did 2.6tons * 3.3*10^10J = 5.3632E29eV which equals 5.3632E23MeV. Then i took that number and divided it by 208MeV/nucleus to get 2.578E21 nuclei. I multiplied that by 1mol/6.23E23 nuclei and 235g/mol to get .9725g of Uranium. Webassign says my answer is within 10% of the correct answer. Am i missing a step?
The 32P nucleus decays into a 32S and electron, but the masses in the table are for neutral atoms. You have 15 electrons on the left to make the 32P neutral, and you need 16 on the right to make 32S neutral. So the electron ejected from the nucleus is needed to give you a neutral atom on the right side.
How do you approach number 8? I don't even know where to start. Am I supposed to use the equation KE_min= (1+(m/M))*absolute value of Q? If so, how do i get the KE if I am looking for mass?
How do you approach number 8? I don't even know where to start. Am I supposed to use the equation KE_min= (1+(m/M))*absolute value of Q? If so, how do i get the KE if I am looking for mass?
Problem #8. I'm sorry I didn't have time to run an example of this in class, but the concept is the same as ones where we had the mass of all the nuclei and found the energy released (Q). We are not concerned with the energy of the incoming particle, so the KEmin=(1_m/M)|Q| relation is not relevant.
Take one of the entries in the table accompanying that problem as an example. 19F(p,alpha)16O with Q=8.114 MeV means that if you hit a 19F nucleus with a proton, it will produce an alpha and a 16O nucleus, and release an additional 8.114 MeV of energy (which will mostly go into the kinetic energy of the alpha). That balances Z and A, but to find the energies from masses, we should work with neutral atoms. That would be 19F+1H --> 4He+16O+8.114 MeV. (Still balances A and Z, and both sides are electrically neutral.) Suppose that you didn’t know the mass of 19F. Get the other masses from appendix B, and convert 8.114 MeV to u: (mass of 19F)+1.007825 = 15.994915+4.002603 + (8.114/931.5). That can be solved for the mass of 19F = 18.998403, which agrees with the value in Appendix B.
Concerning problem 9b, I wasn't sure how to relate the ratio of rads/hour to distance so I found a solution to a similar problem from an online source. This source gave me the equation: I = I(subscript 0)/(4*pi*r^3). Using this equation as the identity for rads/h at some distance I was able to solve the problem, but I am unfamiliar with the notation (thought the equation does look familiar). Is this from a past chapter or something we went over in class I just forgot?
Sorry - didn't directly cover the material for HW #9 in class either. Flux = power / (4 pi r^2) should be familiar from last semester, as the relation between sound intensity and total power, and it's the same thing with radiation - the flux falls as the square of the distance from the source. (Also need to know that rad = rem for gamma rays.)
As I said, I'm sorry that I didn't cover the material for problem #9 in class. It is, however, in the textbook. The use of rads in this context is a bit confusing with respect to radians in angular motion, but a rad is a measure of energy deposition, 0.01 Joule per kilogram. rem (short for Rad Equivalent in huMans) is equivalent to the rad, as long as we're talking about x-ray or gamma rays, as this problem is. Have a look at the post above for discussion of how dose decreases as distance^-2 from the source.
Before anybody asks, I will state unequivocally that there is nothing about biological effects of radiation, units of rad, rem, in the final exam.
17 comments:
I am having trouble with #10. I did 2.6tons * 3.3*10^10J = 5.3632E29eV which equals 5.3632E23MeV. Then i took that number and divided it by 208MeV/nucleus to get 2.578E21 nuclei. I multiplied that by 1mol/6.23E23 nuclei and 235g/mol to get .9725g of Uranium. Webassign says my answer is within 10% of the correct answer. Am i missing a step?
Check your value for Avogadro's number.
on question 7, the energy released does not include the electron's mass, seemingly contrary to both the book and the lecture. whats the deal?
The 32P nucleus decays into a 32S and electron, but the masses in the table are for neutral atoms. You have 15 electrons on the left to make the 32P neutral, and you need 16 on the right to make 32S neutral. So the electron ejected from the nucleus is needed to give you a neutral atom on the right side.
How do you approach number 8? I don't even know where to start. Am I supposed to use the equation KE_min= (1+(m/M))*absolute value of Q? If so, how do i get the KE if I am looking for mass?
How do you approach number 8? I don't even know where to start. Am I supposed to use the equation KE_min= (1+(m/M))*absolute value of Q? If so, how do i get the KE if I am looking for mass?
Please help with number 8. I don't know what equation to use to solve for the mass.
ZOMG I CANNOT FIGURE OUT NUMBER 8. What am I doing wrong? I don't know where to start =(
Help!?
Is webassign still due at 5 pm on Monday?? I have a final during the day so I would appreciate it if the deadline was extended.
Problem #8. I'm sorry I didn't have time to run an example of this in class, but the concept is the same as ones where we had the mass of all the nuclei and found the energy released (Q). We are not concerned with the energy of the incoming particle, so the KEmin=(1_m/M)|Q| relation is not relevant.
Take one of the entries in the table accompanying that problem as an example. 19F(p,alpha)16O with Q=8.114 MeV means that if you hit a 19F nucleus with a proton, it will produce an alpha and a 16O nucleus, and release an additional 8.114 MeV of energy (which will mostly go into the kinetic energy of the alpha). That balances Z and A, but to find the energies from masses, we should work with neutral atoms. That would be 19F+1H --> 4He+16O+8.114 MeV. (Still balances A and Z, and both sides are electrically neutral.)
Suppose that you didn’t know the mass of 19F. Get the other masses from appendix B, and convert 8.114 MeV to u:
(mass of 19F)+1.007825 = 15.994915+4.002603 + (8.114/931.5).
That can be solved for the mass of 19F = 18.998403, which agrees with the value in Appendix B.
Last problem set is due Monday, Dec. 13th, at midnight. Good luck with all your exams!
Prof.
Concerning problem 9b, I wasn't sure how to relate the ratio of rads/hour to distance so I found a solution to a similar problem from an online source. This source gave me the equation: I = I(subscript 0)/(4*pi*r^3).
Using this equation as the identity for rads/h at some distance I was able to solve the problem, but I am unfamiliar with the notation (thought the equation does look familiar). Is this from a past chapter or something we went over in class I just forgot?
Thanks for your help,
Peter J. Davis
Sorry - didn't directly cover the material for HW #9 in class either. Flux = power / (4 pi r^2) should be familiar from last semester, as the relation between sound intensity and total power, and it's the same thing with radiation - the flux falls as the square of the distance from the source. (Also need to know that rad = rem for gamma rays.)
I really do not under stand number 8. Do we need formuls for angular momentum? Please someone help get started.
The last comment was supposed to say number 9.
Sorry
As I said, I'm sorry that I didn't cover the material for problem #9 in class. It is, however, in the textbook. The use of rads in this context is a bit confusing with respect to radians in angular motion, but a rad is a measure of energy deposition, 0.01 Joule per kilogram. rem (short for Rad Equivalent in huMans) is equivalent to the rad, as long as we're talking about x-ray or gamma rays, as this problem is.
Have a look at the post above for discussion of how dose decreases as distance^-2 from the source.
Before anybody asks, I will state unequivocally that there is nothing about biological effects of radiation, units of rad, rem, in the final exam.
Thank you very much!
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