Global Chemistry Lessons
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L.P. #307 S.M.R. Cunupia
L.P. #307 S.M.R. Cunupia
Theory for this syllabus section, U1 M1 SS 1.7 is broken up into 6 small, and easily manageable parts. After going through them, you will be shown how to use each part to answer the only three CAPE past paper questions on the topic. They are:
- 2019 U1 P2 Q1(a) - Link to paper
- 2009 U1 P2 Q1(a) - Link to paper
- 1999 U1 P2 Q4(a)
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Part 1
Each element has a characteristic 'fingerprint' line emission spectrum due to its unique arrangement of electrons/electronic configuration. The simplest of these is produced by hydrogen.
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Part 2
- When sufficient energy is supplied to the atoms of hydrogen, an electron is promoted (excited) from a lower energy level to a higher energy level where it becomes unstable.
- The unstable electron will emit the excess energy as radiation and drop back to the lower energy level. Each line in the emission spectrum occurs due to the transitions of electrons moving from a higher energy level to a lower energy level.
- The energy difference between the higher energy level and lower energy level is fixed. The fixed frequencies of each line provide evidence for discrete energy levels.
- Each line represents emitted radiation of a specific wavelength and frequency. Electronic transitions from higher levels to the level, n = 2 give rise to a series of lines known as the Balmer series.
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Part 3
In summary,
- Electrical or thermal energy is passed through and absorbed by a sample of the element.
- Radiation is emitted at certain wavelengths and frequencies.
- The emission spectrum consists of separate lines at the particular frequencies.
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Part 4
Electron transitions occurring between energy levels associated with the hydrogen emission spectrum are shown in the diagram below.
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Part 5
The three main regions of the hydrogen emission spectrum, as related to series of lines are:
Lyman series < 400 nm - ultraviolet region.
Balmer series 400 nm to 700 nm - visible region.
Paschen series > 700 nm - infrared region.
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Part 6
To calculate the energy, E, of a quantum of radiation with a frequency (ν) of 4.57 × 1014 Hz where h = 4 ×10-13 kJ s mol-1, we use the formula E = hν as illustrated next.
E = h × ν
= 4 ×10-13 kJ s mol-1 × 4.57 × 1014 Hz
= 182.8 kJ mol-1
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How To Answer The Past Paper Questions
2019 U1 P2 Q1(a)(i) - Each point in Part 3 is one mark each.
2019 U1 P2 Q1(a)(ii) - Part 5 says it is the Balmer series.
2019 U1 P2 Q1(a)(iii) - First three points and first sentence in 4th point of Part 2 are one mark each.
2019 U1 P2 Q1(a)(iv) - Lines in Balmer series, only, of Part 4 for full marks.
2009 U1 P2 Q1(a)(i) - Answer is found in Part 1.
2009 U1 P2 Q1(a)(ii) - Draw just Balmer part of diagram in Part 5 and label as required.
2009 U1 P2 Q1(a)(iii) - First, second and fourth points in Part 2 will get you full marks.
2009 U1 P2 Q1(a)(iv) - Part 5 says it is in the visible region.
2009 U1 P2 Q1(a)(v) - Part 6 is the solution.
1999 U1 P2 Q4(a)(i) - Same as 2009 U1 P2 Q1(a)(ii)
1999 U1 P2 Q4(a)(ii) - Same as 2019 U1 P2 Q1(a)(ii)
1999 U1 P2 Q4(a)(iii) - All four points in Part 2 for full marks.