Chapter 6 Flashcards
Describes the theory that James Clerk Maxwell developed.
James Clerk Maxwell developed a mathematical theory to describe light and other forms of radiation in terms of oscillating, or wave like, electric and magnetic fields
Electromagnetic radiation is characterized by its _________ and____________.
Electromagnetic radiation is characterized by its wavelength and frequency
What is the definition of wavelength and its symbol?
Symbolized by greek letter lambda λ
Defined as the distance between a given point on a wave and the corresponding point in the next cycle of the wave.
What is the definition of frequency, its symbol and the unit?
Symbolized by the greek letter nu (ν)
Refers to the number of waves that pass a given point in some unit of time, usually per second.
The unit for frequency, written either as s^(-1) or 1/s and standing for 1 oscillation per second, is called a hertz
To what wavelength and frequency is related to?
Wavelength and frequency are related to the speed (c)
How do we calculate the speed (c) with wavelength and frequency?
c (m/s)= λ(m) X ν(1/s)
electromagnetic radiation consists of ____________ and ______________
electromagnetic radiation consists of oscillating electric and magnetic disturbances
Wavelength-Frequency conversions
The frequency of the radiation used in cell phones covers a range from about 800MHz to 2GHz. (MHz stands for “megahertz”, where 1 MHz= 10^6 1/s; GHz stands for “gigahertz”, where 1 GHz = 10^9 1/s)
The value of the speed of light is: 2.998 X 10^8 m/s
What is the wavelength (in meters) of a cell phone signal operating at 1.12 GHz?
Example 6.1: Chemistry and chemical reactivity 9th edition Student book.
λ= c 2.998 X 10^8 m/s
— = ————————— = 0.268m
v 1.12 X 10^9 1/s
What is the value of the Planck’s constant?
6.6261 x 10^-34 J∙ s
What Max Planck assumed in his experiment?
Assumed that only certain vibrational energies were allowed, and they are related to the frequencies emitted by Planck’s equation:
E (J) = h (J∙s) x ν (s^-1)
Where h=6.6261 x 10^-34 J∙ s
What happens if an oscillator changes from a higher energy to a lower one?
Energy is emitted as electromagnetic radiation, where the difference in energy between the higher and lower energy state is
∆E= E(higher n) - E(lower n) = ∆nhν
What is the Planck’s equation?
E=hν
Where h is the Planck’s constant
v is the frequency
What is the key aspects of Planck’s theory for general chemistry ?
The key aspects of Planck’s theory for general chemistry are that Planck introduced the idea of quantized energies and the equation E=hv, an equation that was to have an important impact on the work of Albert Einstein in explaining another puzzling phenomenon.
Which ideal Planck introduced?
The idea of quantized energies
In which explanation Albert Einstein incorporated Planck’s ideas?
Albert Einstein incorporated Planck’s ideas into an explanation of the photoelectric effect.
What is the Einstein photoelectric effect?
The photoelectric effect is the observation that many metals emit electrons when light shines upon them.
Einstein characterized the Photons, as packets of energy, and stated that the energy of each photon is proportional to the frequency of the radiation as defined by _______________
Einstein characterized the Photons, as packets of energy, and stated that the energy of each photon is proportional to the frequency of the radiation as defined by Plank’s equation.
Energy and chemistry: using Planck’s equation
DVD players use laser that emit red light with a wavelength of 640nm.
What is the energy of one photo of this light? What is the energy of 1.0 mol of photons of red light?
1 photons = 6.022 X 10^23 photons= Avogadro’s number
Example 6.2: Chemistry and chemical reactivity 9th edition Student book.
The wavelength (640nm) expressed in meters id 6.4 X 10^-7 m. The frequency is determined using the wavelength- frequency equation
λ= c 2.998 X 10^8 m/s
— = ————————— = 4.7 X 10^14 1/s
v 6.4 X 10^-7 m
Then the frequency is used to calculate the energy per photon.
E=hν
E per photon = hv=
(6.626 X 10^-34 J∙s/photon) X ( 4.7 X 10^14 1/s)
= 3.1 X 10^-19 J/photon
The energy of a mole of photons is calculated by multiplying the energy per photon by Avogadro’s number
E per mol =
(3.1 X 10^-19 J/photon) X (6.022 X 10^23 photons)
=1.9 X 10^5 J/mol