Question

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the charge on the nucleus and B is 18 2.180 10  J. a) Find the ionization energy of the Be3+ ion in its first excited state in kilojoules per mole. b) Find the wavelength of light given off from the Be3+ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels.

Answer 1

Explanation:

(a) The given data is as follows.

            B = $2.180 \times 10^{-18} J$

            Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

         I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

              = $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

           $8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$  

           = 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

         $\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

   $\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$                

        $\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

                     = $303.7 \times 10^{-10} m$

or,                 = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$.