Answer:
Explanation:
a ) All the electronic levels below n = 3 have same energy so there will not be any evolution of energy in electronic transition from n=3 to n=2 or to n= 1 .
b ) Ionisation energy of ground state = 15 x 10⁻¹⁹ J per electron
= 15 x 10⁻¹⁹ x 6.02 x 10²³ J / mol
= 90 x 10⁴ J/mol
= 900 kJ / mol
c ) the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization will correspond to n = 4 to n = 6
= (11 - 2 ) x 10⁻¹⁹ J
= 9 X 10⁻¹⁹ J
= 9 X 10⁻¹⁹ J / 1.6 X 10⁻¹⁹
= 5.625 eV
= 1244 / 5.625 nm
= 221.155 nm
(a) The highest as well as the minimum frequency will be "n=3 to n=2 to n=1".
(b) The ionization energy will be "900 kJ.mol".
(c) The shortest wavelength will be "221.155 nm".
(a) Because all electronically stages below n=3 have almost the similar energy, there'll be no energy development inside this electronic configuration from n=3 to n=1.
(b)
→ Ionization energy of ground state will be:
= [tex]15\times 10^{-19} \ J/electron[/tex]
= [tex]15\times 10^{-19}\times 6.02\times 10^{23} \ J/mol[/tex]
= [tex]90\times 10^4[/tex]
= [tex]900 \ kJ/mol[/tex]
(c)
→ The shortest wavelength will be:
= [tex](11-2)\times 10^{-19} \ J[/tex]
= [tex]9\times 10^{-19} \ J[/tex]
= [tex]\frac{9\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
= [tex]5.625 \ eV[/tex]
= [tex]221.155 \ nm[/tex]
Thus the above answers are correct.
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