Respuesta :
Explanation:
(a) The given data is as follows.
B = [tex]2.180 \times 10^{-18} J[/tex]
Z = 4 for Be
Now, for the first excited state [tex]n_{f}[/tex] = 2; and [tex]n_{i} = \infinity[/tex] if it is ionized.
Therefore, ionization energy will be calculated as follows.
I.E = [tex]\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})[/tex]
= [tex]8.72 \times 10^{-18} J[/tex]
Converting this energy into kJ/mol as follows.
[tex]8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol[/tex]
= 5249 kJ/mol
Therefore, the ionization energy of the [tex]Be^{3+}[/tex] ion in its first excited state in kilojoules per mole is 5249 kJ/mol.
(b) Change in ionization energy is as follows.
[tex]\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}[/tex]
[tex]\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}[/tex]
[tex]\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}[/tex]
= [tex]303.7 \times 10^{-10} m[/tex]
or, = [tex]303.7^{o}A[/tex]
Therefore, wavelength of light given off from the [tex]Be^{3+}[/tex] ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels [tex]303.7^{o}A[/tex].
