Lab 10 prep

Question 1:

They give you N=100, and say the absolute error is sqrt(N) so absolute error = 10.
By definition we know that absolute error is the relative error times the value (N), so to find relative error you just now divide by your value N(100) and you get .1 . If you complete the same procedure for the next two parts you get 31.62 and .03162.

Question 2:

All I have so far is:
If x is in meters what are the units of the absorption constant ? 1/m

Question 3:

All I have so far is:
In this equation what are the units of lambda if t is in seconds? 1/s

N=N_0*e^(-lambda*x)
we want only 0.01 (or 1%) left so that is what N =
N_0 = 1 (or 100%) which is what we start out with so

0.01/1 = e^-lambda*90 (given this #)
ln (0.01)/90=lambda

Q 3
second part
ln(2)/lambda

did anyone get Question #3 and the rest of Question #2?

I am having trouble with the rest of question 2 as well can someone please help?

rest of Q2:

thickness would have to be .0511685576 m

23.66205057% of rays would be blocked
just use the formula: N=N_o*e^-lambda(x). the new thickness is .003m, lambda is still 90, N_o is 1, and you're looking for N. solve for that, you get that N=.7633794943, but that's the percent that's allowed through, so multiply by 100 for the percentage then subtract from 100 to get 23.66205057

Q3:

units are 1/s
lambda is just ln(2)/half life given in minutes
N=N_o*e^-lambda(t) .. ln(.01)/answer from above question

what is the equation for the half life in terms of( lambda) in Question #3?

ln(2)/lambda


did anyone get the very last question?

how long would it take for 1 percent?

question 3 :


.266 (you do ln2/(2.6min x 60 sec)



17.31 ( ln100/( .266)

in question#3, do anyone know the equation for half life in terms of lambda....this is question #3-part 2?