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Solution: Consider a hydrogen atom in the ... | Clutch Prep
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A hydrogen atom is an atom of a chemical element of hydrogen. Electronally neutral atoms contain one positively charged proton and a negatively charged electron is bound to the nucleus by the Coulomb force. The hydrogen atom represents about 75% of the baryonic mass of the universe.

In everyday life on Earth, isolated hydrogen atoms (called "hydrogen atoms") are extremely rare. In contrast, hydrogen tends to combine with other atoms in the compound, or by itself to form the usual hydrogen (diatomic) gas, H 2 . "Hydrogen atoms" and "hydrogen atoms" in ordinary English use have overlapping, yet different meanings. For example, water molecules contain two hydrogen atoms, but do not contain hydrogen atoms (which will refer to isolated hydrogen atoms).

Attempts to develop a theoretical understanding of hydrogen atoms have been important to the history of quantum mechanics.


Video Hydrogen atom



Isotop

The most abundant isotope, hydrogen-1 , protium , or mild hydrogen , does not contain neutrons and is just a proton and an electron. Protium is stable and forms 99.985% of naturally occurring hydrogen atoms.

Deuterium contains one neutron and one proton. Deuterium is stable and forms 0.0156% of the naturally occurring hydrogen and is used in industrial processes such as nuclear reactors and Nuclear Magnetic Resonance.

Tritium contains two neutrons and one proton and is unstable, decomposed with a half-life of 12.32 years. Because of its short half-life, tritium does not exist in nature except in small quantities.

Higher isotope hydrogen is only made in artificial accelerators and reactors and has half life around the order of 10 -22 seconds.

The formula below applies to the three isotopes of hydrogen, but values ​​slightly different from the Rydberg constant (the correction formula given below) should be used for each hydrogen isotope.

Maps Hydrogen atom



Hydrogen ion

Hydrogen is not found without its electrons in ordinary chemistry (temperature and space pressures), because ionized hydrogen is highly reactive chemically. When ionized hydrogen is written as "H " as in classic acid solvents such as hydrochloric acid, hydronium ions, H 3 O , that is, not atoms single ionized hydrogen. In this case, the acid transfers the proton to H 2 O to form H 3 O .

Ionized hydrogen without its electron, or free proton, is common in interstellar medium, and solar wind.

IBM Discovers a Way of Saving Data on an Atom
src: anonhq.com


Theoretical analysis

The hydrogen atoms have a special significance in quantum mechanics and quantum field theory as the physical system of simple two-body problems that have produced many simple analytical solutions in closed form.

The classic description that failed

Experiments by Ernest Rutherford in 1909 showed the atomic structure to be a solid and positive nucleus with light, a negative charge orbiting around it. This immediately causes problems on how the system can stabilize. Classical electromagnetism has shown that any accelerated charge emits the energy described by the Larmor formula. If electrons are assumed to orbit in perfect circles and radiate energy continuously, electrons will rapidly rotate into the nucleus by falling time:

                       t                       crashed                          ?                     ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ...              a                     Â  <Â> <Â>        ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ,                       Â 3        ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ,          ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ/                Â 4    ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ...         Â                                  0        ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ,                                2        ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ,     ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂ/       ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂï <½                                      ?        1,6         ?                   10                ÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÂÃ, -    Â 11                           s                       {\ displaystyle t _ {\ text {fall}} \ approx {\ frac {a_ {0} ^ {3}} {4r_ {0} ^ { 2} c}} \ approximately 1.6 \ cdot 10 ^ {- 11} {\ text {s}}}  Â

Di                            a                       0                              {\ displaystyle a_ {0}}   adalah radius Bohr dan                            r                       0                              {\ displaystyle r_ {0}}  adalah jari-jari electron klasik. Jika ini benar, semua atom akan segera runtuh, namun atom tampaknya stabil. Selanjutnya, spiral ke dalam akan melepaskan noda frekuensi elektromagnetik sebagai orbit semakin kecil. Sebaliknya, atom diamati hanya memancarkan frekuensi discrit radiasi. Resolved acan terletak pada pengembangan mechanical coat.

Model Bohr-Sommerfeld

In 1913, Niels Bohr obtained the energy and spectral frequencies of the hydrogen atom after making a number of simple assumptions to correct the failed classical model. The assumptions include:

  1. The electrons can only be in a particular circular orbit or stationary , thus having a series of different radii and energy.
  2. Electrons do not emit radiation while in one of these stationary states.
  3. An electron can gain or lose energy by jumping from one discrete orbitals to another.

Bohr suspects that the angular momentum of an electron is quantized by a possible value:

                   L        =        n        ?             {\ displaystyle L = n \ hbar}  di mana                    n        =        1        ,         2        ,         3        ,        .        .        .             {\ displaystyle n = 1,2,3,...}  Â

dan                        ?                  {\ displaystyle \ hbar}    adalah konstanta Planck di atas                         2         ?                  {\ displaystyle 2 \ pi}    . Dia juga menduga bahwa gaya sentripetal yang menjaga elektron di orbitnya disediakan oleh kekuatan Coulomb, dan energi itu dilestarikan. Bohr menurunkan energi setiap orbit atom hidrogen menjadi:

                                   E                         n                              =          -                                                                  m                                     e                                                                 e                                     4                                                                          2                (                4               ?                                ?                                     0                                                                )                                     2                                                                ?                                     2                                                                                                   1                             n                                 2                                                                  {\ displaystyle E_ {n} = - {\ frac {m_ {e} e ^ {4}} {2 (4 \ pi \ epsilon _ {0}) ^ { 2} \ hbar ^ {2}}} {\ frac {1} {n ^ {2}}}}    ,

di mana                            m                      e                              {\ displaystyle m_ {e}}   adalah massa electron,                    e             {\ displaystyle e}   adalah muatan elektron,                            ?                       0                              {\ displaystyle \ epsilon _ {0}}  adalah permitivitas vakum, dan                    n             {\ displaystyle n}  adalah bilangan kuantum (sekarang dikenal sebagai bilangan kuantum utama). Prediksi Bohr cocok dengan experiments from Yang Mengukur rangkaian spectral hydrogen that belongs to the member of the country, and that is what the theory of the Yori menggunakan nilai terkuantisasi.

Untuk                         n          =          1                  {\ displaystyle n = 1}    , nilainya

                                                                                m                                     e                                                                 e                                     4                                                                          2                (                4               ?                                ?                                     0                                                                )                                     2                                                                ?                                     2                                                                           =                                                                  m                                     e                                                                 e                                     4                                                                          8                                 h                                     2                                                                ?                                     0                                                      2                                                                           =          1          R          y          =          13.605                   692                   53          (          30         )                              eV                           {\ displaystyle {\ frac {m_ {e} e ^ {4}} {2 (4 \ pi \ epsilon _ {0}) ^ {2} \ hbar ^ { 2}}} = {\ frac {m _ {\ text {e}} e ^ {4}} {8h ^ {2} \ varepsilon _ {0} ^ {2}}} = 1Ry = 13.605 \; 692 \; 53 (30) \, {\ text {eV}}}   

disebut unit energi Rydberg. Ini terkait dengan konstanta Rydberg                                    R                        ?                                      {\ displaystyle R _ {\ infty}}    fisika atom dengan                         1                              Ry                  ?          h          c                     R                        ?                             .                  {\ displaystyle 1 \, {\ text {Ry}} \ equiv hcR _ {\ infty}.}   

The exact value of the Rydberg constant assumes that the infinite nuclei are related to the electron. For hydrogen-1, hydrogen-2 (deuterium), and hydrogen-3 (tritium), the constants must be slightly modified to use the reduced mass of the system, not just the mass of the electrons. However, since the nucleus is much heavier than electrons, its value is almost the same. The Rydberg constant R M for the hydrogen atom (one electron), R is given by

                              R                      M                         =                                            R                            ?                                                    1                                         m                                   e                                                            /                            M                                     ,             {\ displaystyle R_ {M} = {\ frac {R _ {\ infty}} {1 m {{text {e}}/M}} ,}  Â

di mana                         M                  {\ displaystyle M}    adalah massa inti atom. Untuk hidrogen-1, kuantitas                                    m                         e                                        /                   M         ,                  {\ displaystyle m _ {\ text {e}}/M,}    adalah sekitar 1/1836 (yaitu rasio massa elektron-ke-proton). Untuk deuterium dan tritium, rasionya masing-masing sekitar 1/3670 dan 1/5497. Angka-angka ini, ketika ditambahkan ke 1 dalam penyebut, mewakili koreksi yang sangat kecil dalam nilai R , dan dengan demikian hanya koreksi kecil untuk semua tingkat energi dalam isotop hidrogen yang sesuai.

Masih ada masalah dengan model Bohr:

  1. gagal memprediksi detail-detail spectral lainnya seperti struktur halus dan struktur hyperfine
  2. itu hanya bisa memprediksi tingkat energi dengan acurasi untuk atom elektron tunggal (atom hydrogen-seperti)
  3. nilai prediksi hanya bless untuk                            ?                       2                         ?                 10                       -             5                             {\ displaystyle \ alpha2 \ kira-kira 10 ^ {- 5}}   , di mana                    ?            {\ displaystyle \ alpha}   adalah konstanta struktur-halus.

Much of this deficiency was corrected by Arnold Sommerfeld's modification of the Bohr model. Sommerfeld introduces two additional degrees of freedom that allow electrons to move in elliptical orbits, characterized by eccentricity and declination of the selected axis. It introduces two additional quantum numbers, corresponding to the orbital angular momentum and projection on the selected axis. Thus the diversity of the correct circumstances (except for factor 2 accounting for an unknown electron spin) has been found. Further applying the special theory of relativity to elliptical orbits, Sommerfeld succeeded in lowering the correct expression for the fine structure of the hydrogen spectrum (which occurs exactly as in the most complicated Dirac theory). However, some observed phenomena such as the Zeeman anomaly effect remain unexplained. These problems are solved with the full development of quantum mechanics and Dirac equations. It is often assumed that the SchrÃÆ'¶dinger equation is superior to Bohr-Sommerfeld's theory of describing hydrogen atoms. But this is not the case, since the best results of both approaches are concurrent or very close (exceptional exception is the problem of hydrogen atoms in electric fields and crossed magnets, which can not be solved within the framework of Bohr-Sommerfeld's own theory. ), and their major deficiency results from the absence of electron spin in both theories. The complete failure of Bohr-Sommerfeld's theory to account for the many electron systems (such as helium atoms or hydrogen molecules) that demonstrate his inability to describe quantum phenomena.

Schrödinger's Equation

The Schrödinger equation allows one to calculate the development of a quantum system with time and can provide an appropriate analytical answer for a non-relativistic hydrogen atom.

Wave Function

Hamiltonian atom hydrogen adalah operator energi kinetik radial danaya tarik coulomb ant proton positif under electricity negatif. Menggunakan persamaan Schrö¶dinger yang independen waktu, mengabaikan semua interaksi spin-coupling dan menggunakan massa                    ?        =                m                      e                         M                  /                (                m                      e                              M        )             {\ displaystyle \ mu = m_ {e} M/(m_ {e} M)}   , persamaan ditulis sebagai:

                             (                       -                                                          ?                                      2                                                                   2                  ?                                                                 ?                              2                                      -                                                          e                                      2                                                                    4                  ?                                     ?                                          0                                                        r                                                              )                ?        (        r        ,        ?        ,        ?        )        =        E        ?        (        r        ,        ?        ,        ?        )             {\ displaystyle \ left (- {\ frac {\ hbar ^ 2}} {2 \ mu}} \ nabla2 - {\ frac {e2} {4 \ pi \ epsilon 0} r}} \ right) \ psi (r, \ theta, \ phi) = E \ psi (r, \ theta, \ phi)}  Â

Expanding Laplacian in ball coordinates:

Ini adalah persamaan diferensial parsial yang terpisah yang dapat dipecahkan dalam hal fungsi-fungsi khusus. Fungi gelombang yang dinormalisasi, yang diberikan dalam koordinat bulat adalah:

                           ?                      n            l           m                         (        r        ,        ?        ,        ?        )        =                                                                           (                                                            2                                              n                                                a                                                      0                                                                                *                                                                                                                               )                                                            3                                                                                      (                  n                  -                  l                  -                  1                   )                   !                                                  2                  n                   (                  n                                  l                   )                   !                                                                            e                       -             ?                          /                         2                                ?                      l                                  L                      n             -            l             -            1                                 2            l                     1                         (        ?        )                  Y                      l                               m                         (        ?        ,        ?        )             {\ displaystyle \ psi {n \ ell m} (r, \ vartheta, \ varphi) = {\ sqrt {{\ left ({\ frac { 2} {na_ {0} ^ {*}}} \ right)} ^ {3} {\ frac {(n- \ ell -1)!} {2n (n \ ell)!}}}} E ^ { - \ rho/2} \ rho ^ {\ ell} L_ {n- \ ell -1} ^ 2 \ ell 1} (\ rho) Y _ {\ ell} m (\ vartheta, \ varphi) }  Â

Angka-angka kuantum dapat menambambil nilai-nilai berikut:

                   n        =        1        ,         2        ,         3        ,        ...            {\ displaystyle n = 1,2,3, \ ldots}  Â
                   l        =         0        ,        1        ,         2        ,        ...        ,        n        -        1            {\ displaystyle \ ell = 0,1,2, \ ldots, n-1}  Â
                   m        =        -        l        ,        ...        ,        l        .            {\ displaystyle m = - \ ell, \ ldots, \ ell.}  Â

Selain itu, fungi gelombang ini dinormalkan (yaitu, integral dari modulus persegi mereka same dengan 1) dan orthogonal:

                            ?                       0                               ?                                 r                       2                        d        r                 ?                       0                                 ?                         without             ?        d        ?                 ?                       0                                 2             ?                         d        ?                        ?                      n            l           m                                 *                         (        r        ,        ?        ,        ?        )                ?                                  n               ?                                   l               ?                                   m               ?                                    (        r        ,        ?        ,        ?        )        =         ?        n        ,        l        ,        m                   |                       n           ?              ,                l           ?              ,                m           ?               ?        =                ?                      n                        n               ?                                            ?                      l                        l               ?                                            ?                     m                        m               ?                                    ,             {\ displaystyle \ int _ {0} ^ {\ infty} r2 dr \ int _ {0} ^ {\ pi} \ without \ vartheta d \ vartheta \ int _ {0} ^ 2 \ pi} d \ varphi \; \ psi \ n \ ell m} ^ {*} (r, \ vartheta, \ varphi) \ psi {n '\ ell' m '} (r, \ vartheta, \ varphi) = \ langle n, \ ell , m | n ', \ ell', m '\ rangle = \ delta _ {nn'} \ delta _ {\ ell \ ell '} \ delta mm {},}  Â

di mana                               |                n        ,        l        ,        m         ?             {\ displaystyle | n, \ ell, m \ rangle}   adalah status yang diwakili oleh fungi gelombang                            ?                      n            l           m                              {\ displaystyle \ psi {n \ ell m}}  dalam notasi Dirac, dan                    ?             {\ displaystyle \ delta}  adalah fungsi delta Kronecker.

Fungi gelombang dalam ruang momentum terkait dengan fungsi gelombang dalam ruang posisi melalui transformasi Fourier

                   ?        (        p        ,                ?                      p                         ,                ?                      p                         )        =        (         2        ?        ?                 )                       -             3                          /                         2                         ?                e                       -      Â

Source of the article : Wikipedia

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