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Log Sig Figs Confusion 
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#1
Feb605, 01:55 AM

P: 43

I am rather confused by how the significant figure is suppose to work when you have logarithms in your equation. For example:
Francium87 undergoes beta decay with a half life of 22 min. How much of a 15.00 grams sample remains after 2.0 hours? Well, I used this equation: [tex]logA = log 15.00g \frac{0.301(2.0hours)}{22minutes}[/tex] After solving it, i get: logA = 0.4 > A = 10^0.4 And arrived at an answer of 0.4g This is where i get confused, noticed the bold prints. Am i suppose to round my answers to the sig figs based on the first 1 or the second 1? 


#2
Feb605, 02:56 AM

Sci Advisor
PF Gold
P: 613

I wonder this too, but I think two significant figures is needed, since you were given 15.00 grams of sample.



#3
Feb605, 07:23 AM

P: 1,116

Is that the answer??? 0.4g???
I only ask because I got 0.36g (to 2 significant figures) without using logs. The Bob (2004 ©) 


#4
Feb605, 09:08 AM

Sci Advisor
PF Gold
P: 613

Log Sig Figs Confusion
As The_Bob said, I wouldn't use logarithmbased calculation here; if the half life is 22 mins, two hours (120 mins) have about 5.45 half cycles. When we start with 15.00 grams, here is what I get:
15>7.5>3.75>1.825>0.9375>0.46875 the fifth half cycle is over, and 0.45 half cycle is remaining. 0.50 half cycle is 11 minutes, so 0.45 half cycle equals to 9.9 minutes. If the mass is lowered by 50% in 22 mins, how much of it is lost after 9.9 mins? Calculating with a 15.00 gsample, we get 0.06 grams. If we subtract this from what we've found first, we get 0.40875, which is 0.41 according to two significant figures. Am I correct pals? Edit: If we begin obeying the twofigures expression, the answer is plain 0.4 grams, since 1.825 is equal to 1.83, and so on. 


#5
Feb605, 10:52 AM

P: 43

Well i did get answers similar to you guy's. Except i rounded it to the nearest tenth, so i get 0.4g. I've been responses telling me that i should use the sig fig based on the "15.00" and some other responses telling me to do it the other way. So im rather confused at which is the correct 1 right now.



#6
Feb605, 01:25 PM

P: 1,116

What I did was to also work out the 6th cycle (being 0.234375). The different between cycle 5 and 6 is going to be 0.234375. I then worked out [tex]\frac{5}{11}[/tex] of this, not 0.45, which gives me 0.10653409. Then I subtracted this from the 5th cycle to leave 0.362215 or 0.36 (to 2 significant figures). So have I done something wrong because you are the chemistry guru, I am 16 and you have something different to me. The Bob (2004 ©) 


#7
Feb605, 03:09 PM

P: 43

im confused



#8
Feb605, 05:58 PM

P: 43

ok, i came across another problem with the same thing:
Neptunium231 decays by electron capture. If, after 3.50 hours, a 2.000 gram sample has decaed to only 0.0339g, what is the halflife of 231Np? this sig fig thing is really throwing me off. 


#9
Feb705, 02:07 AM

Sci Advisor
PF Gold
P: 613

Well, if they don't say anything about significant figures like "give your results in three/four significant figures", etc., choose it by selecting the most significant one, in your last example, it is 4 due to 0.0339 grams of remaining mass.
Start from 2.000 grams and divide by 2 for each cycle. You'll see how long does each cycle goes. 


#10
Feb705, 02:32 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

For the original problem :
1. Since you are required to calculate a mass, your accuracy must be the same as the data provided for a quantity with the same units. In this case, that would be the original mass (at t=0) {strictly speaking, multiplied by any dimensionless constant The relative errors in these constants should also propagate. But if you use the power rule for the relative error, you will find that there is no error in the constant term.} The mass is given to be 15.00, which has an accuracy of 4 sig figs. Hence the answer must contain 4 sig figs. 2. This problem must be solved using logarithms. You can not solve it to arbitrary precision using linear interpolation. The Bob's solution is close, but not correct because he used a linear interpolation to the actual function (an exponential). 3. Solve the problem to a reasonably high accuracy (greater than the required accuracy) and finally report the answer with the required number of sig figs. Do not round off constants like log2 to fewer sig figs or this will introduce additional error. 4. You (Dooh, that is) have made a small error in your calculation. You should get logA = 0.46589 . And this will give you the correct answer A = 0.3421 grams. Your equation, however, is correct. 


#11
Feb705, 05:04 PM

P: 1,116

The Bob (2004 ©) P.S. Does that mean that Dooh's other question needs logs as well??? 


#12
Feb705, 05:18 PM

P: 1,116

For the next question that Dooh ask I got down to:
[tex]2^x = 58\frac{338}{339}[/tex] I have forgotten how to solve x but if you can then that should be the answer, in minutes. The Bob (2004 ©) 


#13
Feb705, 05:31 PM

P: 1,116

Just being looking on the good old archives, when PF was very very simplistic, and I have found that x = 5.88 minutes.
[tex]2^x = 58\frac{338}{339}[/tex] [tex]log (2^x) = log 58\frac{338}{339}[/tex] [tex]x log 2 = log 58\frac{338}{339}[/tex] [tex]x = \frac{log 58\frac{338}{339}}{log 2} = 5.88[/tex] The Bob (2004 ©) 


#14
Feb705, 07:26 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

The Bob, x (in your above calculation) does not represent the time in minutes. It should be a dimensionless fraction.
Anyway, we should let Dooh work on this, since the difficulty with the sig figs has been sorted out. 


#15
Feb705, 09:02 PM

P: 43

great, thanks guys. You really helped me through this, now i just gotta make sure i remember all of these for the upcoming test.



#16
Feb805, 01:52 PM

P: 1,116

The Bob (2004 ©) 


#17
Feb805, 08:06 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

x = 5.88 is correct, but x is the number of halflives (ie : t/T). Since t is given, you can then find the halflife, T.



#18
Feb905, 03:59 PM

P: 1,116

The Bob (2004 ©) 


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