Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization energy of the ground state hydrogen atom is 2.17 &#xD7; 10 &#x2212; 11 erg per atom. (Take h = 6.62 &#xD7; 10 &#x2212; 27 e r g &#x2212; s )

Question

Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization energy of the ground state hydrogen atom is    2.17   &amp;#xD7;        10            &amp;#x2212;       11          erg per atom. (Take    h   =   6.62   &amp;#xD7;        10            &amp;#x2212;       27           e   r   g   &amp;#x2212;   s )

tammypierce5kgk · Accepted Answer

Given, Returns from the Bohr orbit of electrons         n     2      =   2   &amp;#xA0;   t   o   &amp;#xA0;        n     1      =   1  state. The hydrogen atom&#039;s ground state ionization energy    =   2.17   &amp;#xD7;        10            &amp;#x2212;       11           e   r   g        /      a   t   o   m  Again, We know Ionisation energy    =   &amp;#x2212;        E     1     i.e ionisation energy is the energy required to remove an electron from ground state, so indirectly we are provided with the value of         E     1    . Also,         E     2      &amp;#x2212;        E     1      =        &amp;#x394;      E (when an electron returns from 2 to 1)   =               h       c                 &amp;#x3BB;            &amp;#x21D2;        &amp;#x3BB;      =               h       c                          E         2              &amp;#x2212;                E         1                 ...(1) Where, h = Planck&#039;s constant               c = speed of light We know,        E     n      &amp;#x221D;               Z       2                 n       2          [Here Z is constant, since we are considering the same atom] Hence,        E     n      &amp;#x221D;        1            n       2                        E       1                 E       n           =               n       2          1     We can say that,        E     n      =               E       1                 n       2          Hence putting n = 2,   &amp;#x21D2;        E     2      =               E       1          4      =               &amp;#x2212;       2.17       &amp;#xD7;                10                    &amp;#x2212;           11                          4     erg/atom   =   &amp;#x2212;   0.5425   &amp;#xD7;        10            &amp;#x2212;       11          erg/atom So,        &amp;#x394;      E   =        E     2      &amp;#x2212;        E     1      =   &amp;#x2212;   0.5425   &amp;#xD7;        10            &amp;#x2212;       11           e   r   g   &amp;#x2212;   (   &amp;#x2212;   2.17   &amp;#xD7;        10            &amp;#x2212;       11           )   =   1.6275   &amp;#xD7;        10            &amp;#x2212;       11           e   r   g  ∴ from equation (1),        &amp;#x3BB;      =               (       6.62       &amp;#xD7;                10                    &amp;#x2212;           27                       )       &amp;#xD7;       (       3       &amp;#xD7;                10                    10                       )                 1.6275       &amp;#xD7;                10                    &amp;#x2212;           11                           =   12.20   &amp;#xD7;        10            &amp;#x2212;       6           c   m   =        1220     o      A

Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization energy of the ground state hydrogen atom is 2.17xx10^(−11) erg per atom. (Take h= 6.62xx10^(−27)erg−s)

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