Disregaurd my incorrect answers, Thank you! The following values are the only al
ID: 555170 • Letter: D
Question
Disregaurd my incorrect answers, Thank you!
The following values are the only allowable energy levels of a hypothetical one-electron atom: E6"-2.0 × 10-19 J Eg =-7.0 × 10-19 J E4-11.0 x 1019 J E315.0 x 10-19 j E2--17.0 x 10-19 J E1--20.0 x 10-19 J (a) If the electron were in the n = 4 level, what would be the highest frequency (and minimum Hint Solution Guided Solution 10 points Print References wavelength) of radiation that could be emitted? Frequency = 1.36 ×10-5 Wavelength = 2.21 ×10-7 (b) What is the ionization energy (in kJ/mol) of the atom in its ground state? x 10-21 kJ/mol 1.8 × (c) If the electron were in the n = 1 level, what would be the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization? 1,105 |×10- nmExplanation / Answer
a) When the electron in n=4 level emitsenergy, the maximum energy(maximum frequency corresponds to transition
4--->1 .
Thus the energy differenc e = E4 -E1
= -11x10-19 -(-20x10-19) J
=9.0x10-19 J
thus frequency = delta E / h
=9.0x10-19 J /6.626x10-34 J
= 1.35x1015 Hz
and the wavelength corrresponding to this energy = hc/E = 2.2x10-7 m
b) The ionisation energy of this is the energy required to remove the electron from the first orbit .
Thus ionisation energy = + 20x10-19 J/ atom
As 1 mole has Avogadro number = 6.626x1023atoms
the ionisation energy per mole = + 20x10-19 J/ atom x6.626x1023atoms
= 1.2 x1013 kJ/mol
c) Without causing ionisation, the electron from n=1 can go to n=6
Thus the energy required = E6-E1 = 18x10-19 J
and the corresponding wavelength = hC/ delta E
= 6.626x10-34 J.sec x 3.0x108m/sec / 18.0x10-19 J
= 110x10-9 m
= 110 nm