What is the probability for an electron of an atom on Earth to lie outside the galaxy?











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In this youtube video it is claimed that electrons orbit their atom's nucleus not in well-known fixed orbits, but within "clouds of probability", i.e., spaces around the nucleus where they can lie with a probability of 95%, called "orbitals".



It is also claimed that the further away one looks for the electron from the nucleus, the more this probability decreases, yet it never reaches 0. The authors of the video conclude that there is a non-zero probability for an atom to have its electron "on the other side of the Universe".



If this is true, then there must be a portion of all atoms on Earth whose electron lies outside the Milky Way. Which portion of atoms has this property?










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  • 10




    What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
    – FGSUZ
    17 hours ago






  • 1




    @FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
    – DanielSank
    16 hours ago










  • @DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
    – FGSUZ
    16 hours ago






  • 1




    @FGSUZ, yes, I really think it answers the question.
    – DanielSank
    16 hours ago










  • Okay then, I just did.
    – FGSUZ
    6 hours ago















up vote
4
down vote

favorite
1












In this youtube video it is claimed that electrons orbit their atom's nucleus not in well-known fixed orbits, but within "clouds of probability", i.e., spaces around the nucleus where they can lie with a probability of 95%, called "orbitals".



It is also claimed that the further away one looks for the electron from the nucleus, the more this probability decreases, yet it never reaches 0. The authors of the video conclude that there is a non-zero probability for an atom to have its electron "on the other side of the Universe".



If this is true, then there must be a portion of all atoms on Earth whose electron lies outside the Milky Way. Which portion of atoms has this property?










share|cite|improve this question




















  • 10




    What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
    – FGSUZ
    17 hours ago






  • 1




    @FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
    – DanielSank
    16 hours ago










  • @DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
    – FGSUZ
    16 hours ago






  • 1




    @FGSUZ, yes, I really think it answers the question.
    – DanielSank
    16 hours ago










  • Okay then, I just did.
    – FGSUZ
    6 hours ago













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





In this youtube video it is claimed that electrons orbit their atom's nucleus not in well-known fixed orbits, but within "clouds of probability", i.e., spaces around the nucleus where they can lie with a probability of 95%, called "orbitals".



It is also claimed that the further away one looks for the electron from the nucleus, the more this probability decreases, yet it never reaches 0. The authors of the video conclude that there is a non-zero probability for an atom to have its electron "on the other side of the Universe".



If this is true, then there must be a portion of all atoms on Earth whose electron lies outside the Milky Way. Which portion of atoms has this property?










share|cite|improve this question















In this youtube video it is claimed that electrons orbit their atom's nucleus not in well-known fixed orbits, but within "clouds of probability", i.e., spaces around the nucleus where they can lie with a probability of 95%, called "orbitals".



It is also claimed that the further away one looks for the electron from the nucleus, the more this probability decreases, yet it never reaches 0. The authors of the video conclude that there is a non-zero probability for an atom to have its electron "on the other side of the Universe".



If this is true, then there must be a portion of all atoms on Earth whose electron lies outside the Milky Way. Which portion of atoms has this property?







quantum-mechanics electrons wavefunction probability estimation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









knzhou

40.1k11113194




40.1k11113194










asked 18 hours ago









Klangen

1244




1244








  • 10




    What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
    – FGSUZ
    17 hours ago






  • 1




    @FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
    – DanielSank
    16 hours ago










  • @DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
    – FGSUZ
    16 hours ago






  • 1




    @FGSUZ, yes, I really think it answers the question.
    – DanielSank
    16 hours ago










  • Okay then, I just did.
    – FGSUZ
    6 hours ago














  • 10




    What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
    – FGSUZ
    17 hours ago






  • 1




    @FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
    – DanielSank
    16 hours ago










  • @DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
    – FGSUZ
    16 hours ago






  • 1




    @FGSUZ, yes, I really think it answers the question.
    – DanielSank
    16 hours ago










  • Okay then, I just did.
    – FGSUZ
    6 hours ago








10




10




What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
– FGSUZ
17 hours ago




What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. From here to outside of the milky way there are so many perturbations that the model would fail. Does it really make sense to calculate such probability if anything can capture that electron much more easily?
– FGSUZ
17 hours ago




1




1




@FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
– DanielSank
16 hours ago




@FGSUZ please post that as an answer, as it is the most correct thing written on this page so far. The $10^{-31}$ estimates posted below are silly for exactly the reason explained in your comment.
– DanielSank
16 hours ago












@DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
– FGSUZ
16 hours ago




@DanielSank Thank you, I'm flattered, but do you really think it's an answer? I don't know if it actually "answers" the question..
– FGSUZ
16 hours ago




1




1




@FGSUZ, yes, I really think it answers the question.
– DanielSank
16 hours ago




@FGSUZ, yes, I really think it answers the question.
– DanielSank
16 hours ago












Okay then, I just did.
– FGSUZ
6 hours ago




Okay then, I just did.
– FGSUZ
6 hours ago










3 Answers
3






active

oldest

votes

















up vote
14
down vote



accepted










The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,



$$
a_0 = 5.29times 10^{-11} ~{rm m}
$$



For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)



$$
P(r) approx e^{-2r/a_0}
$$



Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the galaxy from an atom on Earth is around



$$
P sim e^{-10^{32}}
$$



that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$.






share|cite|improve this answer























  • I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
    – el duderino
    1 hour ago










  • @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
    – Ryan
    5 mins ago


















up vote
11
down vote













What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.



Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.



Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.



So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.






share|cite|improve this answer




























    up vote
    5
    down vote













    Given a single electron, what is the probability that it is found outside the Milky Way?
    We can estimate it using the ground state wave function of the Hydrogen atom,
    $$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
    where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
    $|psi|^2$ is the probability density, integrating gives
    $$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
    Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
    $$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$



    This number is so small, it is hardly possible to grasp really how small it is.
    There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
    $$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
    which doesn't even make any dent.






    share|cite|improve this answer





















    • Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
      – JEB
      15 hours ago













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    14
    down vote



    accepted










    The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,



    $$
    a_0 = 5.29times 10^{-11} ~{rm m}
    $$



    For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)



    $$
    P(r) approx e^{-2r/a_0}
    $$



    Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the galaxy from an atom on Earth is around



    $$
    P sim e^{-10^{32}}
    $$



    that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$.






    share|cite|improve this answer























    • I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
      – el duderino
      1 hour ago










    • @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
      – Ryan
      5 mins ago















    up vote
    14
    down vote



    accepted










    The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,



    $$
    a_0 = 5.29times 10^{-11} ~{rm m}
    $$



    For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)



    $$
    P(r) approx e^{-2r/a_0}
    $$



    Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the galaxy from an atom on Earth is around



    $$
    P sim e^{-10^{32}}
    $$



    that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$.






    share|cite|improve this answer























    • I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
      – el duderino
      1 hour ago










    • @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
      – Ryan
      5 mins ago













    up vote
    14
    down vote



    accepted







    up vote
    14
    down vote



    accepted






    The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,



    $$
    a_0 = 5.29times 10^{-11} ~{rm m}
    $$



    For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)



    $$
    P(r) approx e^{-2r/a_0}
    $$



    Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the galaxy from an atom on Earth is around



    $$
    P sim e^{-10^{32}}
    $$



    that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$.






    share|cite|improve this answer














    The quantity you should consider first is the Bohr radius, this tells you an idea of the relevant atomic scales,



    $$
    a_0 = 5.29times 10^{-11} ~{rm m}
    $$



    For hydrogen (the most abundant element), in its ground state, the probability of finding an electron beyond a distance $r$ from the center looks something like (for $r gg a_0$)



    $$
    P(r) approx e^{-2r/a_0}
    $$



    Now let's plug in some numbers. The virial radius of the Milky Way is around $200 ~{rm kpc} approx 6times 10^{21}~{rm m}$, so the probability of finding an electron outside the galaxy from an atom on Earth is around



    $$
    P sim e^{-10^{32}}
    $$



    that's ... pretty low. But you don't need to go that far to show this effect, the probability that an electron of an atom in your foot is found in your hand is $sim 10^{-10^{10}}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago









    Nayuki

    13826




    13826










    answered 17 hours ago









    caverac

    4,6132620




    4,6132620












    • I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
      – el duderino
      1 hour ago










    • @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
      – Ryan
      5 mins ago


















    • I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
      – el duderino
      1 hour ago










    • @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
      – Ryan
      5 mins ago
















    I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
    – el duderino
    1 hour ago




    I think it's also important to note that this prediction utilizes the Schrodinger equation, which is non-relativistic. My math isn't up to scratch enough to properly interpret the Dirac equation solution of the hydrogen atom, but my hunch is that it might make such a big jump impossible to avoid causality violations.
    – el duderino
    1 hour ago












    @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
    – Ryan
    5 mins ago




    @elduderino Arguably the "jump" doesn't transmit any information faster than the speed of light (since the transmitter can't force it deterministically, and the receiver can't determine where it came from), so causality isn't necessarily violated.
    – Ryan
    5 mins ago










    up vote
    11
    down vote













    What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.



    Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.



    Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.



    So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.






    share|cite|improve this answer

























      up vote
      11
      down vote













      What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.



      Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.



      Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.



      So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.






      share|cite|improve this answer























        up vote
        11
        down vote










        up vote
        11
        down vote









        What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.



        Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.



        Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.



        So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.






        share|cite|improve this answer












        What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results.



        Take into account that the models are based on hypothesis, which are easily violated. For example, spherical symmetry, which allows finding the solution in the hydrogen atom (or more accurately, the Coulomb's potential in QM). Reality is never like that, but we can say that "it is close enough" if the atom is far enough from other objects.



        Nevertheless, from here to outside of the milky way there are so many perturbations that the model would just fail. You can say that there's a level $n=1324791$, but there are so many particles out there that the effect of your atom is absolutely beaten by ANY other.



        So, does it really make sense to calculate such probability if anything can capture that electron much more easily? I don't think so.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        FGSUZ

        3,5032522




        3,5032522






















            up vote
            5
            down vote













            Given a single electron, what is the probability that it is found outside the Milky Way?
            We can estimate it using the ground state wave function of the Hydrogen atom,
            $$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
            where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
            $|psi|^2$ is the probability density, integrating gives
            $$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
            Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
            $$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$



            This number is so small, it is hardly possible to grasp really how small it is.
            There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
            $$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
            which doesn't even make any dent.






            share|cite|improve this answer





















            • Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
              – JEB
              15 hours ago

















            up vote
            5
            down vote













            Given a single electron, what is the probability that it is found outside the Milky Way?
            We can estimate it using the ground state wave function of the Hydrogen atom,
            $$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
            where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
            $|psi|^2$ is the probability density, integrating gives
            $$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
            Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
            $$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$



            This number is so small, it is hardly possible to grasp really how small it is.
            There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
            $$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
            which doesn't even make any dent.






            share|cite|improve this answer





















            • Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
              – JEB
              15 hours ago















            up vote
            5
            down vote










            up vote
            5
            down vote









            Given a single electron, what is the probability that it is found outside the Milky Way?
            We can estimate it using the ground state wave function of the Hydrogen atom,
            $$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
            where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
            $|psi|^2$ is the probability density, integrating gives
            $$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
            Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
            $$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$



            This number is so small, it is hardly possible to grasp really how small it is.
            There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
            $$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
            which doesn't even make any dent.






            share|cite|improve this answer












            Given a single electron, what is the probability that it is found outside the Milky Way?
            We can estimate it using the ground state wave function of the Hydrogen atom,
            $$ psi_{100} = frac{1}{sqrt{pi a_0^3}} e^{-r/a_0} , $$
            where $a_0 approx 5*10^{-11}, m$ is the Bohr radius.
            $|psi|^2$ is the probability density, integrating gives
            $$ p_1 = int_R^infty |psi_{100}|^2 4pi r^2, dr = frac{e^{-2R/a_0}(a_0^2 + 2a_0 R + 2R^2)}{a_0^2} . $$
            Plugging in $R approx 5*10^{20}, m$ the radius of the Milky Way, we get
            $$ p_1 approx exp(-2*10^{31}) approx 10^{-10^{31}} . $$



            This number is so small, it is hardly possible to grasp really how small it is.
            There are a lot of electrons in the Earth - about $N = 10^{51}$ - but the number of electrons is utterly tiny compared to these odds. The chance that any electron is found outside the milky way is
            $$ p = 1 - (1 - p_1)^N approx N p_1 = 10^{51} , cdot , 10^{-10^{31}} $$
            which doesn't even make any dent.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 17 hours ago









            Noiralef

            3,7861927




            3,7861927












            • Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
              – JEB
              15 hours ago




















            • Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
              – JEB
              15 hours ago


















            Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
            – JEB
            15 hours ago






            Since $10^{-51} approx e^{-117}$, can I estimate that the electron furthest from its nucleus in the all the Earth is about 117 Bohr radii away, on average, at any given time. Roughly?
            – JEB
            15 hours ago




















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