A2 Physics - Nuclear Physics

a) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in 10000 deflected by more than 90º)
b) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm.

Part a is hard to explain in words.
Part b I have an answer of 5*10^-12m, or is it 5*10^-14m?

(edited 10 years ago)

Reply 1

Cora Lindsay

10

a) think about someone striking a snooker ball (the alpha) so that it hits a cannonball (the nucleus). If it hits off-centre, it will deflect through a fairly small angle, but if it hits head-on, it will deflect back the way it came. The fact that only a few alphas deflect through more than 90 degrees tells you that very few of them hit the massive nucleus head on, and that means the nucleus must be very small.

5*10^-14m looks a much mroe sensible number

Reply 2

hello calum

10

Original post by Math12345

a) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in 10000 deflected by more than 90º)
b) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm.

Part a is hard to explain in words.
Part b I have an answer of 5*10^-12m, or is it 5*10^-14m?

Just to clarify, you know it is $\frac{1}{10000}n$ and not $\frac{1}{10000n}$
A couple of questions on that part of the textbooks has come up before so i just want to make sure the books havent tricked you too

by a given atom oops :colondollar:

(edited 10 years ago)

Reply 3

Math12345

OP

14

My teacher said the answer was 3.55*10^-14. The probability is (1/10000n). The n is the confusing bit, because you had to divide the thickness by the diameter of the atom to find n.

Thanks for the help anyways.

(edited 10 years ago)

Reply 4

hello calum

10

Original post by Math12345

My teacher said the answer was 3.55*10^-14. The probability is (1/10000n). The n is the confusing width, because you had to divide the thickness by the diameter of the atom to find n.

Thanks for the help anyways.

Oh sorry about that, I didn't read the "for a given atom" part :colondollar:

You can just explain it by saying that: for each layer that is added, there are more atoms in the path of the beam, so the probablity of the alpha particle hitting the given nucleus is inversley proportional the the number of layers.

Reply 5

Cora Lindsay

10

Original post by hello calum

You can just explain it by saying that: for each layer that is added, there are more atoms in the path of the beam, so the probablity of the alpha particle hitting the given nucleus is inversley proportional the the number of layers.

True, depending on the relationship between the incident alphas and the crystal structure of the gold foil. Gold is face centred cubic, so If the alphas are perpendicular to one of the faces all the atoms in deeper layers are just hiding behind those in the first two layers and the probability of collision won't increase. This is a special case though.

Reply 6

Billyearley

The formula is d^2 = D^2 / (10,000 x n)

d is the diameter of the nucleus, D is the diameter of the atom, you find n by thickness/atom diameter then plug in numbers. It gives me 3.54 x 10^-14 :smile:

I kept getting 5x10^-9 but then saw that formula :smile:

hope it helps :smile:

just realised sed the posts a year and a half old, but hope it helps anyone cheating on homework instead then :wink:

Reply 7

Alen.m

11

i respectfully disagree,
the question mentioned quite clear that the probability 1/10000n is equal to the ratio of the cross-sectional area of the nucleus to that of the atom which means 1/10000n =pi x r^2/pi x R^2 while (r is the radius of the nucleus and R is radios of the atom) and n=thickness divided by diameter of the atom. if you found r in the equation above you can then times it by two which will give you the answer of 3.54 x 10^-14

A2 Physics - Nuclear Physics

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