- #1

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ok i used the equation En=

__k^2 x e^4 me__.

__Z^2__

2xh^2 h^2

i am getting n^2= -9.07x10^21

i know this is wrong and am getting quite frustrated

if some on could help me it would be great

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- Thread starter fatboy12341
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- #1

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ok i used the equation En=

2xh^2 h^2

i am getting n^2= -9.07x10^21

i know this is wrong and am getting quite frustrated

if some on could help me it would be great

- #2

quasar987

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- #3

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You know that the distance of closest approach is of the order of the radius of the Gold atom. The formula for the radius is (if you are aware of the derivation):fatboy12341 said:

ok i used the equation En=k^2 x e^4 me.Z^2

2xh^2 h^2

i am getting n^2= -9.07x10^21

i know this is wrong and am getting quite frustrated

if some on could help me it would be great

[tex]R = \frac{1}{4\pi \epsilon_0}\frac{2Ze^2}{mv^2}[/tex]

If you can see, the KE of the atom has been incorporated in this equation. You just have to substitute the required quantities to obtain the distance of closest approach. You can take it from here I hope...

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kinetic energy = electrical potential

Ke = that's given

Elec Pot = KQq/r

rearrange for r and make sure that k has the correct units.

- #5

quasar987

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I had forgotten about this thread. And since the OP appears to have abandonned it as well, I hereby claim possesion of it!

I don't see how the problem has a solution if we are not given the distance 'd' between the gold sheet and the alpha particle source because, if we take a coordinate system where the origin is on the gold sheet (x-axis perpenducular to it) the potential energy of the "alpha particle-fixed gold nucleus" system is

[tex]U(x)=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_{\alpha}}[/tex]

The energy is a constant of the motion, hence for any two positions [itex]x_1[/itex] and [itex]x_2[/itex] of the alpha particle, we have

[tex]U(x_1)+K(x_1)=U(x_2)+K(x_2)[/itex]

Taking [itex]x_1=d[/itex], the position just as the alpha particle leaves the source (where [itex]K=K_0=7.7[/itex] MeV) and [itex]x_2=x_m[/itex], the minimum distance between the alpha particle and the gold nucleus (where K=0), we have

[tex]\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{d}+K_0=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_m}[/tex]

Solving for x_m requires the knowledge of the distance 'd' btw the source and the gold sheet.

I don't see how the problem has a solution if we are not given the distance 'd' between the gold sheet and the alpha particle source because, if we take a coordinate system where the origin is on the gold sheet (x-axis perpenducular to it) the potential energy of the "alpha particle-fixed gold nucleus" system is

[tex]U(x)=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_{\alpha}}[/tex]

The energy is a constant of the motion, hence for any two positions [itex]x_1[/itex] and [itex]x_2[/itex] of the alpha particle, we have

[tex]U(x_1)+K(x_1)=U(x_2)+K(x_2)[/itex]

Taking [itex]x_1=d[/itex], the position just as the alpha particle leaves the source (where [itex]K=K_0=7.7[/itex] MeV) and [itex]x_2=x_m[/itex], the minimum distance between the alpha particle and the gold nucleus (where K=0), we have

[tex]\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{d}+K_0=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_m}[/tex]

Solving for x_m requires the knowledge of the distance 'd' btw the source and the gold sheet.

Last edited:

- #6

nrqed

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quasar987 said:I had forgotten about this thread. And since the OP appears to have abandonned it as well, I hereby claim possesion of it!

I don't see how the problem has a solution if we are not given the distance 'd' between the gold sheet and the alpha particle source because, if we take a coordinate system where the origin is on the gold sheet (x-axis perpenducular to it) the potential energy of the "alpha particle-fixed gold nucleus" system is

[tex]U(x)=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_{\alpha}}[/tex]

The energy is a constant of the motion, hence for any two positions [itex]x_1[/itex] and [itex]x_2[/itex] of the alpha particle, we have

[tex]U(x_1)+K(x_1)=U(x_2)+K(x_2)[/itex]

Taking [itex]x_1=d[/itex], the position just as the alpha particle leaves the source (where [itex]K=K_0=7.7[/itex] MeV) and [itex]x_2=x_m[/itex], the minimum distance between the alpha particle and the gold nucleus (where K=0), we have

[tex]\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{d}+K_0=\frac{1}{4\pi \epsilon_0}{\frac{q_{\alpha}Q_{Au}}{x_m}[/tex]

Solving for x_m requires the knowledge of the distance 'd' btw the source and the gold sheet.

I thought that in that kind of problem, it was implicitly assumed that the initial distance can be approximated to be infinite. In practice, if the energy is given at, say, a couple of meters from the gold nucleus, the initial potential energy will be completely negligible compared to th efinal potential energy (since the distance of approach is *much* smaller than a few meters).

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