Help calculating the the activity of a radioactive sample

AI Thread Summary
To calculate the activity of potassium-40 (K40) in the body, the mass of K40 is determined to be 1.68 x 10^-5 kg. The number of K40 atoms is calculated using Avogadro's number, resulting in approximately 2.5284 x 10^17 atoms. The activity is then computed using the formula R = λN, where λ is the decay constant derived from the half-life of K40. However, the initial calculations yield an incorrect activity value of 3.88 x 10^-45 Bq instead of the expected 4200 Bq. The discussion highlights the need for accurate conversion and application of formulas to achieve the correct activity measurement.
pearlblack
Messages
2
Reaction score
0
Help! calculating the the activity of a radioactive sample!

1. find activity R



2. the average 70 KF males contain 140 gram of potassium. of this 0.012% is potassium 40. the half life of potassium 40 is 1.2 billion years. so how many K40 decay per second and what is the activity of K40 in the body??



3. so my solution here is..
140g = 0.14kg
and mass of K40 = 0.14 * 0.012/100 = 1.68 * 10^-5kg
how many decay per second = (1.68*10^-5)/2 / 3600*24*365*1.2*10^12 = 2.22*10^-25kg
then i use R = \lambda * N = 0.653N/half life
= 3.88 * 10^-45 Bq
but this answer is wrong! because the correct one is 4200 Bq
so help me please! thx
 
Physics news on Phys.org


Okay, you've got the mass of K40 = 1.68*10^-5 kg. You need to calculate how many K40 atoms there are by multiplying by Avogadro's number and dividing by the molar mass (which is 40).
 


sorry i don't get it..
do i just
1.68*10^-5kg * 6.02*10^23/40 = 2.5284*10^17
and then sub this into the equation R=0.653N/half-life??
but the answer i get from this is 4.36*10^-3

which i think there is still something wrong in ym calculation...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top