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# OCR Physics A AS Forces+Motion/Electrons & Photons revision thread watch

1. nice one!!!! thanks

Isotope U 235(nucleon no) 92(atomic number)

half life 0.75*10^9

number at time after 4.5*10^9 years????

how can I do this?
3. also

describe briefly an absorbtion experiment to distinguish between alpha, beta and gamma radiations

What kind of experiment and sketch could I draw?
4. OCR FFE

Can someone explain the moles business.

SO there are 6.02*10^23 particles in a mole of any substance??

eg: one mole of gold would have this many particles and one mole of silver the same?

say they said 0.1kg of gold how would I find number of moles?
5. (Original post by chats)
OCR FFE

Can someone explain the moles business.

SO there are 6.02*10^23 particles in a mole of any substance??

eg: one mole of gold would have this many particles and one mole of silver the same?

say they said 0.1kg of gold how would I find number of moles?
yeah a mole always has the same number of particles no matter what its of but the weight of a mole is different. the weight of one mole of any substance is equal to its relative atomic mass (grams) so to work out number moles in 0.1kg
100g/mass of gold
6. For the first question, you need some kind of indication of the initial number of nuclei - whether a mass, the number of moles, or the actual number. The numbers you have given (apart from the time) are all constants... they will remain the same no matter how much sample is left so cannot be used to deduce the initial value.

Your second question - use a Geiger counter and place various materials between it and the source (eg. paper, aluminium and lead).

Oldak.
7. Thanks for the second part, thats what I thought.

For the first part the question is

Rutherford suggested that the isotopes of uranium had equal numbers of atoms, when the earth was formed. The half lives of two isotopes are shown

Uranium n-235, a-92 half life 0.75*10^9

Uranium n-238 a-92 half life 4.5*10^9

calculate ratio: number of U-235/ number of U-238

at a time 4.5*10^9 years after the Earth was formed
8. oh thanks!

When would the molar gas constant come into it (8.31)
9. u use it when u use the idea gas equation pv=nrt u have to work out n the number of moles and then u plug it in and r=8.31 the gas constanty thing
10. u use it when u use the idea gas equation pv=nrt u have to work out n the number of moles and then u plug it in and r=8.31 the gas constanty thing
11. You could invent a convenient initial value of number of nuclei (eg. 3*10^15), one which will not be reduced to negligability by the time. Then use λt½=ln2 to work out the decay constant of each. Apply each to N=N0e-λt. This will give you the numbers of nuclei left, based upon the initial value - the ratio between the two will be the same whatever the initial value you picked. Divide U-235/U-238 and you have the ratio... I think.

I am in a rush, so have not really thought of a way in which to do it without value invention - anyone got any ideas?

Oldak.
12. Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 468x986
\Large

^{235} U: t_{\frac{1}{2}} = 0.75 \times 10^9 \\ \, \\

^{238} U: t_{\frac{1}{2}} = 4.5 \times 10^9 \\ \, \\

\lambda = ln2 /t_{\frac{1}{2}} \\ \, \\

\lambda_{235} = ln2/(0.75 \times 10^9) = 9.242 \times 10^{-10} \\ \, \\

\lambda_{238} = ln2/(4.5 \times 10^9) = 1.54 \times 10^{-10} \\ \, \\

N = N_0 e^{-\lambda t} \\ \, \\

N_{235} = N_0 e^{-9.242 \times 10^{-10} 4.5 \times 10^9} \\ \, \\

N_{238} = N_0 e^{1.54 \times 10^{-10} 4.5 \times 10^9} \\ \, \\

\frac{N_{235}}{N_{238}} = \frac{N_0 e^{(-9.242 \times 10^{-10} 4.5 \times 10^9)}}{N_0 e^{(-1.54 \times 10^{-10} 4.5 \times 10^9})} \\ \, \\

\frac{N_{235}}{N_{238}} = 0.0312

hmm there's probably a shorter way.
Is this from a past paper?
13. niceee! thanks
14. Oh yeah here's a shorter way:

at t=4.5*10^9,

for U-235, number of halflives occured= (4.5*10^9)/(0.75*10^9) = 6 halflives
for U-238, number of halflives occured = 1

so N_235/N_238 = 1/(2^6)/(1/2) = 0.03125

I think that's correct.
15. yeah endeavours right, i've done that same past paper and thats how i did it and it checks out with the mark scheme.
U-235 has halved
U-238 has gone down by 1/(2^6)
16. (Original post by g33k b0y)
yeah endeavours right, i've done that same past paper and thats how i did it and it checks out with the mark scheme.
U-235 has halved
U-238 has gone down by 1/(2^6)
17. Thanks both

It was 21 June 2002

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