Ch 243 Quiz
Page 1 of 1 • Share
Ch 243 Quiz
Question 1: ((ln(factor))/(decay constant))/31536000
The number in the denominator is the amount of seconds in a year.
The number in the denominator is the amount of seconds in a year.
helping Guest
Re: Ch 243 Quiz
Question 2:
decays per day/(e^("thousand years ago value"*365000*(.69315/(half life * 365))))
Answer is negative.
decays per day/(e^("thousand years ago value"*365000*(.69315/(half life * 365))))
Answer is negative.
helping Guest
Re: Ch 243 Quiz
Question 3:
(9*10^9*(# of protons)*(1.6*10^19)^2)/(distance)
Answer is positive.
Question 4:
((9*10^9*2*(# of protons)*(1.6*10^19)^2)/distance)/(1.6*10^13)
Answer is negative.
Good luck!
(9*10^9*(# of protons)*(1.6*10^19)^2)/(distance)
Answer is positive.
Question 4:
((9*10^9*2*(# of protons)*(1.6*10^19)^2)/distance)/(1.6*10^13)
Answer is negative.
Good luck!
helping Guest
Re: Ch 243 Quiz
Number 4 is exactly like number 3 except that you multiply by two in the numerator and then divide the whole answer by 1.6E13.
helping Guest
Question 2
I am having problems figuring out number 2. I am using the formula provided above but it doesn't seem to work for me.
The half life of radioactive C14 nuclei is 5730 years. If the activity of a sample today is 24.61 decays per day, what was its activity 2.426 thousand years ago (in decays per day) (sheet 23,23') ? Indicate with a negative (positive) sign whether the number of radioactive nuclei as a function of time obeys the same (a different) exponential decay law as the activity as a function of time.
24.61/(e^(2.426*365000*9.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
The half life of radioactive C14 nuclei is 5730 years. If the activity of a sample today is 24.61 decays per day, what was its activity 2.426 thousand years ago (in decays per day) (sheet 23,23') ? Indicate with a negative (positive) sign whether the number of radioactive nuclei as a function of time obeys the same (a different) exponential decay law as the activity as a function of time.
24.61/(e^(2.426*365000*9.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
Guest786 Guest
question 2 correction
this is what i did
24.61/(e^(2.426*365000*(.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
24.61/(e^(2.426*365000*(.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
Guest786 Guest
Re: Ch 243 Quiz
this is what i did
24.61/(e^(2.426*365000*(.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
ur missing a () sign after e^
it should be 24.61/(e^(2.426*365000*(.69315/(5730*365))))
24.61/(e^(2.426*365000*(.69315/(5730*365)))) and I get my answer as 18.3510....what am i doing wrong?
ur missing a () sign after e^
it should be 24.61/(e^(2.426*365000*(.69315/(5730*365))))
guest123 Guest
Number 2
Hey guys, I got the other 3 but for some strange reason im stuck on number 2. I tried to follow all the appropriate steps but keep getting an incorrect answer on mapleTA . This is my question...
2) The half life of radioactive C14 nuclei is 5730 years. If the activity of a sample today is 20.85 decays per day, what was its activity 8.613 thousand years ago (in decays per day) (sheet 23,23') ? Indicate with a negative (positive) sign whether the number of radioactive nuclei as a function of time obeys the same (a different) exponential decay law as the activity as a function of time
This is what i did...
20.85/ (e^(8.163 * 365000 * (.69315/ (5730* 365))))= 55.97
mapleTA says im wrong. Maybe one of you guys can help me .
Obviously im making a retarded mistake somewhere and I cant seem to figure out where lol so pleaseee. Will be eternally grateful!!!
2) The half life of radioactive C14 nuclei is 5730 years. If the activity of a sample today is 20.85 decays per day, what was its activity 8.613 thousand years ago (in decays per day) (sheet 23,23') ? Indicate with a negative (positive) sign whether the number of radioactive nuclei as a function of time obeys the same (a different) exponential decay law as the activity as a function of time
This is what i did...
20.85/ (e^(8.163 * 365000 * (.69315/ (5730* 365))))= 55.97
mapleTA says im wrong. Maybe one of you guys can help me .
Obviously im making a retarded mistake somewhere and I cant seem to figure out where lol so pleaseee. Will be eternally grateful!!!
Money23 Guest
Re: Ch 243 Quiz
ditto, can someone please explain the mathematical formulation that brings one to the answer for number one?
zZtoP Guest
Re: Ch 243 Quiz
You use the formula on Sheet 23 N(t)=Noe^lambdat
divide No over so you get N/No which is the same thing as the factor of decrease or increase.
N/No = e^lamdat
ln(N/No) = lamdat
(ln(N/No)/lamda) = t
and once you solve for t you gotta convert it to years which is the big number you divide by.
divide No over so you get N/No which is the same thing as the factor of decrease or increase.
N/No = e^lamdat
ln(N/No) = lamdat
(ln(N/No)/lamda) = t
and once you solve for t you gotta convert it to years which is the big number you divide by.
PhyPhy Guest
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

