1.1 EM radition and atomic spectra Flashcards
EM radiation dual nature
can act as both a particle and as a wave
EM radiation in increasing wavelength/decreasing frequency
gamma rays → x rays → ultraviolet → visible light → infrared → microwaves → radiowaves
gamma rays → x rays → ultraviolet → visible light → infrared → microwaves → radiowaves
EM radiation in increasing wavelength/decreasing frequency
frequency and wavelength calculations
- c = f × λ
- c is speed of light (3 x 10^8 m s^-1)
- f is frequency (Hz or m^-1)
- λ is wavelength (m/nm)
- wavenumber = 1 / λ
- λ is wavelength (m/nm)
- f = 1 / T
- T is time (seconds)
what is the frequency of light with a wavelength of 690nm?
f = c / λ
= 3 x 10^8 / 690 x 10^-9 = 4.35 x 10^14 Hz
when do we multiply wavelength and what do we multiply by
multiply wavelength by 10^-9 if wavelength is given in nm
what is the wavelength and frequency of alkyne (wavenumber = 3300 cm^-1)?
wavenumber = 1 / λ
= 1/3300 = 0.000003m
f = c / λ
= 3 x 10^8 / 3.03 x 10^-6 = 9.9 x 10^13 s^-1
energy, frequency, and wavelength calculations
- E = L x h x f
- E is energy (kJ/mol^-1)
- L is Avogadro’s constant (6.023 x 10^23 mol^-1)
- h is Plancks constant (6.63 x 10^-34 J/s)
- f is frequency (Hz)
- E = L x h x c / λ
- E is energy (kJ/mol^-1)
- L is Avogadro’s constant (6.023 x 10^23 mol^-1)
- h is Plancks constant (6.63 x 10^-34 J/s)
- c is the speed of light (3 x 10^8)
- λ is wavelength (m/nm)
what provides evidence for energy levels of atoms
spectrum that is made up of a series of lines at discrete energy levels when light energy is emitted by an atom
relationship between energy of a photon and frequency
it is proportional
- high frequency photon = high energy
- low frequency photon = low energy
two types of atomic spectra
atomic emission spectra
atomic absorption spectra
the concentration of an element within a sample is related to what in electroscopy
the intensity of light emitted or absorbed
atomic absorption spectra and ions
the higher the concentration of metal ions in a solution, the higher the intensity of radiation absorbed
colour of a highly ionic solution that absorbs light only in the UV region
colourless
how can the quantity of an element be determined with electroscopy
from the intensity of the radiation emitted
emission spectroscopy steps
- high temperatures used to excite the electrons
- excited electrons are promoted from their ground state to a higher energy levels
- electrons drop to lower energy levels and emit photons of light
- each photons is equal in energy to the difference between energy levels
- these photons of light can be gathered through a slit and pasted through a prism, which splits the wavelength
what does atomic emission spectroscopy measure
wavelength of emitted radiation by measuring the intensity of light at different wavelengths
what measures wavelength of emitted radiation by measuring the intensity of light at different wavelengths
atomic emission spectroscopy
what does atomic emission spectroscopy produce
a black backdrop with coloured lines, each representing one transition of an electron from a higher energy level back down to ground state
what produces a black backdrop with coloured lines, each representing one transition of an electron from a higher energy level back down to ground state
atomic emission spectroscopy
why does an atomic emission spectra have multiple lines
due to many different energy levels or many different electron transitions
atomic emission spectra and ions
as the concentration of an ionic solution decreases, radiation transmitted increases in intensity
atomic absorption spectra steps
- sample is atomised and EM radiation is directed at the sample
- radiation is absorbed as electrons are promoted to higher energy levels by absorption of specific photons
- each photons is equal in energy to the difference between energy levels
- the light can be passed through a slit, then a prism
what does atomic absorption measure
wavelength of absorbed radiation is measured by measuring how the intensity of absorbed light varies with wavelength
what does atomic absorption spectra produce
ainbow backdrop with missing lines. each line represents an electron being excited from ground state to a higher energy level
calculate the wavelength of a photon of light with the frequency 5 x10^14 s^-1
6 x 10^-7
calculate the wavelength of a photon of light with the frequency 3.2 x 10^14 s^-1
9.4 x 10^-7 m
calculate the frequency of a photon of light with the wavelength 4.26 x 10^-7m
7 x 10^14 s^-1
calculate the energy of one mole of photons in kJ mol for photons with a wavelength of 400nm
300 kJ mol
how is a line produced on an emission spectra
an electron is excited and promoted to a higher energy level
when it falls back down to ground state, energy is emitted
calculate the energy of one mole of photons in kJ mol for photons with a wavelength of 650nm
184 kJ mol
what causes electrons to become excited in atomic emission spectra
heat
A sample of strontium was exposed to electromagnetic radiation with a frequency of 3·08 × 1017 s−1
Calculate the energy, in J, of this electromagnetic radiation
2.04x10-16 J
when do you use L in frequency calculations
if the answer is asking for a mole/in mol units
The concentration of ethanol in vodka can be determined by reacting the ethanol with excess acidified potassium dichromate solution.
20·0 cm3 of vodka was transferred to a 1 litre volumetric flask and made up to the mark with deionised water. 1·0 cm3 of the diluted vodka was pipetted into a conical flask. 25·0cm3 of 0·010moll−1 acidified potassium dichromate was added to the conical flask. The conical flask was then stoppered and warmed until the reaction was complete.
3C2H5OH(aq) + 2Cr2O72−(aq) + 16H+(aq) ↓ 3CH3COOH(aq) + 4Cr3+(aq) + 11H2O(l)
It was found that 1·65 × 10−4 moles of dichromate ions were left unreacted
Calculate the concentration of ethanol, in mol l−1, in the undiluted vodka
Total Dichromate in flask:
no. of mol = volume x concentration = 0.025litres x 0.010 mol l-1 = 2.5x10-4mol
Dichromate left at end of reaction:
no. of mol = 1.65x10-4mol (in question)
Dichromate which reacted with ethanol:
no. of mol = 2.5x10-4 mol – 1.65x10-4mol = 8.5x10-5mol
6.4 mol l-1
why do you make a reactant be in excess
to ensure all of it is reacted
back titration
A known quantity of either an acid or base is reacted with an unknown quantity of some chemical. How much leftover acid or base remaining can be determined via titration and so how much of it reacted with the other chemical can be worked out to determine the chemical’s concentration.
