S6EN Chemistry Homework
Electronic structure of the atom
Information given:
1.0 ev = 1.6 x 10-19 J
Planck’s constant: h = 6.6 x 10-34 J.s
Speed of light: c = 3.0 x 108 m.s-1
Question 1
The diagram below shows the electron energy levels of a hydrogen atom.
a)Mark the ground state on the diagram[1]
The light from a hydrogen electrical discharge tube contains electromagnetic radiation that is blue in colour. This electromagnetic radiation can be separated by a prism to form a line emission spectrum.
b)Describe the physical appearance of a hydrogen line emission spectrum[2]
c)Explain what happens to give rise to the physical appearance (note that a description of electronic transitions is required). [3]
The electron in a hydrogen atom is in the n = 2 state. When it drops to the ground state a photon is emitted.
d)What is the wavelength of the photon in nm? [3]
e)In what region of the electromagnetic spectrum does this emission occur?[1]
Question 2
The energy level diagram below shows some of the outer energy levels of the mercury atom.
a)What is the ionisation energy of mercury in:
i)eV[1]
ii)Joules[2]
b)An electron has been excited to the -1.6 eV level. On the copy of the diagram, show all the possible ways it can return to the ground state. (Note that if you need more space, reproduce the diagram on your test sheet). [2]
c)An experiment is set up whereby a vacuum tube containing gaseous mercury is radiated with monochromatic light (i.e. photons that are of a single wavelength). If all the electrons of the gaseous mercury atoms are in the ground state to begin with, deduce and justify whether the photon energy will be absorbed, if:
i)The photons have an energy of 6.0 eV[2]
ii)The photons have a wavelength of 140 nm.[3]
iii)For i) and ii) above, if energy absorption does occur, and the excited electron falls back down to the -3.7 eV energy level, calculate the wavelength of the emitted photons. [3]
d) Explain what would happen if an electron in the ground state of a mercury atom would absorb a photon of 12.0 eV. [1]
Answers
Question 1
d) E = -3.40 – (-13.6) = 10.2 eV
1.0 ev = 1.6 x 10-19 J 10.2 eV = 1.6 x 10-18 J
e) It is in the UV region of the em-spectrum
Question 2
a)
i) E = Efinal- Eground = 0 – (-10.4) = 10.4 eV
ii) 1.0 ev = 1.6 x 10-19 J 10.4 ev = 1.7 x 10-18 J
b) Route 1: From -1.6 ev to -3.7 ev followed by -3.7 eV to -5.5 eV followed by -5.5 eV to -10.4 eV
Route 4: From -1.6 ev to -5.5 ev followed followed by -5.5 eV to -10.4 eV
Route 3: from -1.6 eV to -5.5 eV followed by -5.5 eV to -10.4 eV
Route 4: from -1.6 eV to -10.4 eV
c)
i) No, light will not be emitted as the energy of the photon does not have the correct quanta i.e. amount of energy to allow for an electronic transition to occur. That is:
E = Efinal- Eground
E = -5.5 – (-10.4) = 4.9 eV
E = -3.7 – (-10.4) = 6.7 eV
E = -1.6 – (-10.4) = 8.8 eV
ii) 1.0 ev = 1.6 x 10-19 J 1.4 x 10-18 J = 8.8 eV
Yes, the photon has the correct amount of energy to result in an energy transition from the ground state to the -1.6 eV state.
iii)
E = -1.6 – (-3.7) = 2.1 eV
1.0 ev = 1.6 x 10-19 J 2.1 eV = 3.4 x 10-19 J
d) Ionisation i.e. enough energy to completely remove the electron from the outermost shell.