Respuesta :
Explanation:
[tex]E=\frac{h\times c}{\lambda}[/tex]
where,
E = energy of photon = ?
h = Planck's constant = [tex]6.634\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda [/tex] = Wavelength of the radiation
As we can see that wavelength and energy of the radiation are inversely related to each other.So with increase in energy wavelength decreases and vice versa.
Given: The energy of different of different energy levels in an hypothetical model of atom;
[tex]E_6 =-2\times 10^{-19} J[/tex]
[tex]E_5 = -7\times 10^{-19} J[/tex]
[tex]E_4 = -11 \times 10^{-19} J[/tex]
[tex]E_3 = -15\times 10^{-19} J[/tex]
[tex]E_2 = -17\times 10^{-19} J[/tex]
[tex]E_1 = -20\times 10^{-19} J[/tex]
a)If the electron is initially in the n = 4 level. Then maximum difference of energy in energy levels will corresponds to shortest wavelength.
So, [tex]E=E_4-E_1[/tex]
[tex]E=-11 \times 10^{-19} J-(-20 \times 10^{-19} J)=9 \times 10^{-19} J[/tex]
[tex]\lambda =\frac{hc}{E}[/tex]
[tex]=\frac{6.634\times 10^{-34}Js\times 3\times 10^8m/s}{9 \times 10^{-19} J}[/tex]
[tex]\lambda = 2.211\times 10^{-7}m\approx 2\times 10^{-7} m[/tex]
[tex]2\times 10^{-7} m[/tex]is the shortest wavelength of radiation that could be emitted.
b) Ionization energy of the atom in its first excited state.
[tex]E'=E_{\infty }-E_2[/tex]
[tex]E'=0-(-17\times 10^{-19} J)=17\times 10^{-19} J[/tex]
Ionization energy of the 1 mole :
[tex]I.E.=\frac{17\times 10^{-19} J}{\frac{1}{6.022\times 10^{23}} mol}[/tex]
[tex]=1,203,740 J/mol=1203.740 kJ/mol\approx 1.2\times 10^3 kJ/mol[/tex]
[tex] 1.2\times 10^3 kJ/mol[/tex] is the ionization energy of the atom in its first excited state.
