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AS radioactivity question!!
A smoke detector contains a small radioactive source. A typical source contains 1.2 x 10^-8 g of americium-241, which has a half-life of 432 years.
a) Show that the decay constant of americium-241 is approximately 5 x 10^-11 s^-1. (where does the 0.69 come from?!)
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei.
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
a) Show that the decay constant of americium-241 is approximately 5 x 10^-11 s^-1. (where does the 0.69 come from?!)
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei.
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
You have N(t) = N(0) exp (-Rt)
where R is decay constant.
We have information about the half life, so using the definition of half life as the time taken for the sample to decay to half the original amount:
1/2 = 1 x exp[-Rt] where t is now the half life. Therefore taking logs we get:
R = ln(1/2) / - t
or R= ln(2) / t
ln(2) is 0.69
Nuclear radii are of the order of a femtometre (10^-12).
where R is decay constant.
We have information about the half life, so using the definition of half life as the time taken for the sample to decay to half the original amount:
1/2 = 1 x exp[-Rt] where t is now the half life. Therefore taking logs we get:
R = ln(1/2) / - t
or R= ln(2) / t
ln(2) is 0.69
Nuclear radii are of the order of a femtometre (10^-12).
a) Show that the decay constant of americium-241 is approximately 5 x 10^-11 s^-1. (where does the 0.69 come from?!)
turn the 431 years into seconds, 431*365.25*24*60*60
divide 0.69 (ln2) by this number and u get just over 5.0*10^-11.
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei
(6 x 10^23/241)*1.2 x 10^-8=3.0x10^13
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
A=(lamda)N
so multiply the 2 numbers u got, i got about 1500Bq
i thought one protons width was the order of a fento-metre? and i thought one fento-metre was 10^-15? Im not sure about this though, will someone check my answers.
turn the 431 years into seconds, 431*365.25*24*60*60
divide 0.69 (ln2) by this number and u get just over 5.0*10^-11.
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei
(6 x 10^23/241)*1.2 x 10^-8=3.0x10^13
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
A=(lamda)N
so multiply the 2 numbers u got, i got about 1500Bq
i thought one protons width was the order of a fento-metre? and i thought one fento-metre was 10^-15? Im not sure about this though, will someone check my answers.
*girlie*
(where does the 0.69 come from?!)
The ln button is on your calculator. Well, should be anyway.
Just press ln then 2.
*girlie*
A smoke detector contains a small radioactive source. A typical source contains 1.2 x 10^-8 g of americium-241, which has a half-life of 432 years.
a) Show that the decay constant of americium-241 is approximately 5 x 10^-11 s^-1. (where does the 0.69 come from?!)
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei.
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
a) Show that the decay constant of americium-241 is approximately 5 x 10^-11 s^-1. (where does the 0.69 come from?!)
b) Calculate the number of nuclei in 1.2 x 10^-8 g of americium-241, given that 241 g contains 6 x 10^23 nuclei.
c) Hence calculate the activity of 1.2 x 10^-8 g of americium-241.
This is where the 0.69 comes from, but you don't need to know it for A level physics (infact I only did radioactivity at A2), and it's using A level maths, so ignore it if you don't do maths.
after one halflife..
Unparseable latex formula:
\large 0.5N_0 \ = \ N_0 e^{-\lambda t} \\ \, \\[br]\frac{1}{2} \ = \ e^{-\lambda t} \\ \, \\[br]ln\frac{1}{2} \ = \ -\lambda t \\ \, \\[br]-\frac{ ln\frac{1}{2} } {\lambda} \ = \ t \\ \, \\[br]t \ = \ \frac{ln2}{\lambda}
ln2 = 0.693
a) You've probably been given the formula
decay constant = 0.693/half life
so decay constant = 0.693/(432*365*24*60*60)
= 5*10-11 s-1
b) N = nNa where N = number of particles; n = number of moles; Na = Avagadro's constant
n = mass/Mr
N = (mass/Mr)Na
N = (1.2*10-8 /241)6*1023
N = 3.0*1013
c)
A = λN
A = 5*10-11 * 3.0*1013
A = 1500 Bq
endeavour
This is where the 0.69 comes from, but you don't need to know it for A level physics (infact I only did radioactivity at A2), and it's using A level maths, so ignore it if you don't do maths.
Quick, close your eyes!!!!!
Well a carbon-12 nuclei will have 12 nucleons, so ~ 12*10-15m ~ 10-14m.
*girlie*
The diameter of a carbon nucleus is approximately 10^n m. Suggest a value for n.
Are we just meant to know this?
Are we just meant to know this?
Yeah you need to know the diameter of nucleus and diameter of an atom.
Atom = 10^-10
Nucleus = 10^-15
I did this Jan 2005 paper today aswell! I lost my marks on mainly on qs 6 and 8.
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