What happens to the first ionization potential when a hydrogen-like atom captures a particle?












2












$begingroup$


This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:




The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?




My answer is that it may either increase or decrease depending on the charge of the captured particle.



However, the correct answer according to my book is that the ionization potential increases.



How do I arrive at this solution?






physical-chemistry ionization-energy atomic-structure







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
    $endgroup$
    – Soumik Das
    2 hours ago


















2












$begingroup$


This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:




The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?




My answer is that it may either increase or decrease depending on the charge of the captured particle.



However, the correct answer according to my book is that the ionization potential increases.



How do I arrive at this solution?






physical-chemistry ionization-energy atomic-structure







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
    $endgroup$
    – Soumik Das
    2 hours ago
















2












2








2





$begingroup$


This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:




The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?




My answer is that it may either increase or decrease depending on the charge of the captured particle.



However, the correct answer according to my book is that the ionization potential increases.



How do I arrive at this solution?






physical-chemistry ionization-energy atomic-structure







share|improve this question









$endgroup$




This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:




The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?




My answer is that it may either increase or decrease depending on the charge of the captured particle.



However, the correct answer according to my book is that the ionization potential increases.



How do I arrive at this solution?






physical-chemistry ionization-energy atomic-structure




physical-chemistry ionization-energy atomic-structure






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









user69284user69284

434




434








  • 1




    $begingroup$
    Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
    $endgroup$
    – Soumik Das
    2 hours ago
















  • 1




    $begingroup$
    Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
    $endgroup$
    – Soumik Das
    2 hours ago










1




1




$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago






$begingroup$
Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = - frac{me^4Z^2}{8 epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.
$endgroup$
– Soumik Das
2 hours ago












1 Answer
1






active

oldest

votes


















2












$begingroup$

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:



$$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$



the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:



$$ a_0 = frac{hbar^2}{me^2} tag{2} $$



The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:



$$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$



If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.



Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:



$$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$



the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.



A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?






share|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "431"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f110000%2fwhat-happens-to-the-first-ionization-potential-when-a-hydrogen-like-atom-capture%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:



    $$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$



    the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:



    $$ a_0 = frac{hbar^2}{me^2} tag{2} $$



    The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:



    $$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$



    If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.



    Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:



    $$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$



    the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.



    A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:



      $$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$



      the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:



      $$ a_0 = frac{hbar^2}{me^2} tag{2} $$



      The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:



      $$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$



      If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.



      Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:



      $$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$



      the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.



      A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:



        $$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$



        the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:



        $$ a_0 = frac{hbar^2}{me^2} tag{2} $$



        The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:



        $$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$



        If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.



        Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:



        $$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$



        the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.



        A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?






        share|improve this answer









        $endgroup$



        When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:



        $$ psi_{1s} = frac{2}{a_o^{3/2}} e^{-r/a_0} tag{1} $$



        the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:



        $$ a_0 = frac{hbar^2}{me^2} tag{2} $$



        The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:



        $$ m = frac{m_e m_p}{m_e + m_p} tag{3} $$



        If we take the limit of $m_p to infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.



        Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:



        $$ I = frac{me^4}{8 epsilon_0 ^2 n^2h^2} tag{4} $$



        the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.



        A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        John RennieJohn Rennie

        1,422918




        1,422918






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Chemistry Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f110000%2fwhat-happens-to-the-first-ionization-potential-when-a-hydrogen-like-atom-capture%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
            Read more