# Ch 24-3 Quiz

## Ch 24-3 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 24-3 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 24-3 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 24-3 Quiz

Number 4 is exactly like number 3 except that you multiply by two in the numerator and then divide the whole answer by 1.6E-13.

**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 24-3 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 24-3 Quiz

ditto, can someone please explain the mathematical formulation that brings one to the answer for number one?

**zZtoP**- Guest

## Re: Ch 24-3 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

Similar topics

» Manners, Etiquette & Politeness Quiz

» QUIZ!!! More Little Cherubs, Imps and Assorted Devils to Identify!

» FUNNY QUIZ - The 'Politico Palm Puzzler'

» QUIZ - 'A Hands-Journey Around the World'!

» FAMOUS SIMIAN LINES - Which famous persons have a 'simian line'?

» QUIZ!!! More Little Cherubs, Imps and Assorted Devils to Identify!

» FUNNY QUIZ - The 'Politico Palm Puzzler'

» QUIZ - 'A Hands-Journey Around the World'!

» FAMOUS SIMIAN LINES - Which famous persons have a 'simian line'?

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum