Why scattered electron waves from different parts of a nucleus interfere destructively?How to prove the...
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Why scattered electron waves from different parts of a nucleus interfere destructively?
How to prove the equivalence of two definitions of the scattering cross sectionWhy is number of collisions equal to number of scattered particles times number of scatterers?Form Factor in Rutherford ScatteringWhat does it mean that the Rutherford's cross section is infinite?How to reconcile infinite cross section of resonances with cross section formula from quantum mechanics?
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We know that,
$sigma (theta)=sigma _M (theta){F(q)}^2$, where $sigma _M (theta)$ is the scattering cross section for a point nucleus while $sigma (theta)$ is the observed scattering cross section from a nucleus of a finite size.Here, F(q) is the form factor.
$sigma _M (theta)=frac{Ze^2}{8pi epsilon _0 E}frac{cos^2 theta /2}{sin^4 theta/2}$
In a book, it's given that the form factor F(q) is lower than one as the scattered electron waves from different parts of the nucleus interfere destructively. If you could help me to understand why they interfere destructively, I'd appreciate that.
nuclear-physics scattering scattering-cross-section
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
$endgroup$
add a comment |
$begingroup$
We know that,
$sigma (theta)=sigma _M (theta){F(q)}^2$, where $sigma _M (theta)$ is the scattering cross section for a point nucleus while $sigma (theta)$ is the observed scattering cross section from a nucleus of a finite size.Here, F(q) is the form factor.
$sigma _M (theta)=frac{Ze^2}{8pi epsilon _0 E}frac{cos^2 theta /2}{sin^4 theta/2}$
In a book, it's given that the form factor F(q) is lower than one as the scattered electron waves from different parts of the nucleus interfere destructively. If you could help me to understand why they interfere destructively, I'd appreciate that.
nuclear-physics scattering scattering-cross-section
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
$endgroup$
add a comment |
$begingroup$
We know that,
$sigma (theta)=sigma _M (theta){F(q)}^2$, where $sigma _M (theta)$ is the scattering cross section for a point nucleus while $sigma (theta)$ is the observed scattering cross section from a nucleus of a finite size.Here, F(q) is the form factor.
$sigma _M (theta)=frac{Ze^2}{8pi epsilon _0 E}frac{cos^2 theta /2}{sin^4 theta/2}$
In a book, it's given that the form factor F(q) is lower than one as the scattered electron waves from different parts of the nucleus interfere destructively. If you could help me to understand why they interfere destructively, I'd appreciate that.
nuclear-physics scattering scattering-cross-section
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
$endgroup$
We know that,
$sigma (theta)=sigma _M (theta){F(q)}^2$, where $sigma _M (theta)$ is the scattering cross section for a point nucleus while $sigma (theta)$ is the observed scattering cross section from a nucleus of a finite size.Here, F(q) is the form factor.
$sigma _M (theta)=frac{Ze^2}{8pi epsilon _0 E}frac{cos^2 theta /2}{sin^4 theta/2}$
In a book, it's given that the form factor F(q) is lower than one as the scattered electron waves from different parts of the nucleus interfere destructively. If you could help me to understand why they interfere destructively, I'd appreciate that.
nuclear-physics scattering scattering-cross-section
nuclear-physics scattering scattering-cross-section
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
asked 8 hours ago
Debasish DhalDebasish Dhal
293 bronze badges
293 bronze badges
add a comment |
add a comment |
1 Answer
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One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.
Interference effects always appear in wave solutions of wave equations but the context is very important.
As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.
It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.
Single electrons seem random, the interference shows the wave nature of the underlying equation.
The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.
So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves.
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
$endgroup$
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.
Interference effects always appear in wave solutions of wave equations but the context is very important.
As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.
It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.
Single electrons seem random, the interference shows the wave nature of the underlying equation.
The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.
So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves.
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
$endgroup$
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
add a comment |
$begingroup$
One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.
Interference effects always appear in wave solutions of wave equations but the context is very important.
As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.
It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.
Single electrons seem random, the interference shows the wave nature of the underlying equation.
The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.
So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves.
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
$endgroup$
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
add a comment |
$begingroup$
One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.
Interference effects always appear in wave solutions of wave equations but the context is very important.
As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.
It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.
Single electrons seem random, the interference shows the wave nature of the underlying equation.
The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.
So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves.
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
$endgroup$
One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.
Interference effects always appear in wave solutions of wave equations but the context is very important.
As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.
It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.
Single electrons seem random, the interference shows the wave nature of the underlying equation.
The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.
So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves.
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
answered 8 hours ago
anna vanna v
167k8 gold badges161 silver badges467 bronze badges
167k8 gold badges161 silver badges467 bronze badges
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
add a comment |
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points.
$endgroup$
– Debasish Dhal
6 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
$begingroup$
I have not at hand the definitions, but usually there are normalizations , and I am sure that the mathematics is correct in defining the crossections given a form factor formalism. I am just trying to clear up that interference does not happen for individual particles, but shows up in the probability distributions derived from the wavefunction. The double slit is given as an easy example.
$endgroup$
– anna v
4 hours ago
add a comment |
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