Chem Exam 3

What is meant by the term electromagnetic radiation and what are its (physical) properties?

Waves and light are considered electromagnetic radiations. Electromagnetic radiations physical properties are dependant on wavelengths and frequencies. This is known as the electromagnetic spectrum. It includes gamma rays, x rays, far ultra-violet, near ultra violet, visible, near infrared, far infrared, microwaves, radio waves, and radar.

A wave is constituted by its frequency and wavelength. Wave length is represented by lambda. Lambda is the exactly the distance between to points, or peaks on the wave.

Wavelength is calculated using nanometers. (1 nm = 10-9 m) See example, 7.2 pg. 267--Heather, this is why you use 10-9!

Frequency is how many times a wave passes a point during a second. See figure 7.4 for a simple explanation of wave length and frequency, and to see how they relate.

Important Calculation:

When asked to calculate wavelength use below formula:
Lambda (distance)=c (speed of light constant 3.00 x 108 m/s)/v (frequency)

When asked to calculate frequency use below formula:
V (frequency)=c (speed of light constant)/lambda (distance)












Our desire is to learn something more about the atom and atomic structure. In chapter 2 we learned about Dalton's Atomic Theory (the atomic nature of matter). The nuclear theory of the atom had us to understand that the atom has a compact nucleus and is largely empty space containing the electrons outside of the nucleus. Describe what is happening in figures 7.1 and 7.2. What kind(s) of information are figures 7.1 and 7.2 providing? What kind(s) of information are figures 7.3-7.5 providing? In figure 7.4 the top wave has a frequency of 2 Hertz. What does that mean and how would I be able to determine that from the figure?
Further study: problems 7.35-7.38

In figure 7.1 a “wire loop” is placed into a flame. Depending on the placement of the wire loop the light emitted from the flame is different. In the first picture the loop is placed horizontal and close to the gas apparatus. The metal element releases the atoms in gas form and show the element and its particular color. For example, the first picture in 7.1 is lithium, which produced a red color, and a blue color for the gas. In the second picture, the wire loop was moved vertically and higher up, producing a yellow color, and again a blue color from the gas. Yellow emission of atoms is an indication of the element Na, sodium. In figure 7.2 each element is placed onto a chart, showing the color that corresponded to the visible light emitted by the elements atoms. The wavelength is also determined for each element and its numerical characteristics in nanometers.
In figure 7.3 the illustration is of a water wave ripple. This figure shows Lambda (wavelength), with identical peaks of the ripples, exactly the same space from each other.
In figured 7.4 the relationship between frequency and wavelength are correlated. See first discussion post for a more detailed explanation. You can determine the top wavelength in figure 7.4 as 2 Hz because of the two equal peaks frequency. The bottom example have 4 identical peaks in one second, giving it a frequency of 4 Hz.
Practice Problem 7.35
Since the question is asking for wavelength use this equation Lambda (distance)=c (speed of light constant 3.00 x 108 m/s)/v (frequency)
3.00x108 m/s/1.365 x106/s = 219.6 m (because seconds cancel)
Practice Problem 7.37
V (frequency)=c (speed of light constant)/lambda (distance)
V=3.00x108 m/s/478x 10-9= 6.27 x 1014/s

Problem 1
The Black Body Problem was an experiment Max Planck did. He heated solids and watched them change color as the temperature was increased and decreased. For example, at 750 degrees Celsius a solid metal object glows red, and at 1200 degrees Celsius a metal solid glows white. I think this concept of heating solids to observe a color is known as blackbody radiation.

Planck’s experiment created a theory that the atoms in a solid vibrate, and only have a particular vibration (E).
Equation:
E=nhv
E(atom vibration)=n(quantum number, or size)h(Planck’s constant – 6.63 x 10-37)v(frequency)

In the attached figure, another theory (Rayleigh-Jeans Law) is when a solid is experimented with at low temperatures. The wavelength doesn’t pass the horizontal axis, which means it slows down, and is not visible. There is no vibration.

Problem 2
Photoelectric effect is a theory from Einstein. He used Planck’s data about quantum theory and expanded on it. Einstein theorized that light (from any source) is made up of photons, and the photons eject electrons from objects (metals) it hits. In order to see a color the amount of electrons ejected need to be over a certain number, also known as “threshold value.”


Light behaves like matter because of the electromagnetic energy (photon) in the light. When photons hit an object the energy is absorbed in the object, and the photon is gone. Einstein’s equation shows the relationship.

Equation:
E=hv
E(vibration)=h(Planck’s constant – 6.63 x 10-37)v(frequency)


In figure 7.6 electromagnetic energy or photons, are being shined on a mental object. The metal object is ejecting the electrons and they are being caught by the positive wire connected to the metal. This is causing the ammeter in the battery to move because of the capturing of the ejected electrons from the photons (or photoelectric energy).









An energy of 2.0 x 102 kJ/mol is required to cause a cesium atom on a metal surface to lose an electron. What is the quanta of energy per atom? Calculate the longest possible wavelength of light that can ionize a cesium atom. In what region of the electromagnetic spectrum is this radiation found?


Problem one
Since the equation is E=hv, I think you can change the equation to get quanta into v=E/n

2.0 x 102 / 6.63 x 10-35 = 3.02 x 1036 (not sure what units)

Longest possible wavelength
This is the longest as possible, because the speed of light is constant (?).

Radio Waves (?)

Heather –
Did you get this?


You are an engineer designing a switch that works by the photoelectric effect. The metal you wish to use in your device requires 6.7 x 10-19 J/atom to remove an electron. Will the switch work if the light falling on the metal has a wavelength of 540 nm or greater? Why or why not?

No the switch will NOT work if the lights wavelength is 540 nm. The quanta frequency would need to be more than double.

Why because of Mathematical Explanation:

Lambda=3.00 x 108 / 540 x 10-9 = 5.56 x 1014

E= (6.63 x 10-34)(5.56 x 1014) = 3.69 x10-19 (required 6.7 x 10-19 J/atom)









Problem 7.49

WOW I think I got this!

(-RH/25)-(-RH/9)=-9RH+25RH/225=16RH/225=hv

v=16RH/225h=16/225(2.179x10-18)/(6.626x10-34)=2.34 x 1014

Problem 7.50

Am I right?

(-RH/16)-(-RH/9)=-9RH+16RH/144=7RH/144=hv

v=7RH/144h=7/144(2.179x10-18)/(6.626x10-34)=1.60 x 1014

Problem 7.51

I know this is right!!!

(-RH/4)-(-RH/1)=-1RH+4RH/4=3RH/4=hv

v=3RH/4h=3/4(2.179x10-18)/(6.626x10-34)=1.21 x 10-7 or 121 nm, Near Ultra Violet

Problem 7.52

Is this right?

(-RH/25)-(-RH/16)=-16RH+25RH/400=9RH/400=hv

v=9RH/400h=9/400(2.179x10-18)/(6.626x10-34)= 7.40 x 1013, Near Infrared


I do not understand what you are looking for***Using figure 7.11 and Example 7.4, assign these bands to electron transitions within the hydrogen atom****

Exercise 7.5
I need help with this. I don’t understand







Part A. de Broglie Relation
Exercise 7.6
6.63 x 10-34/2.19 x 106 = 3.03 x 10-40 or 303 pm (??)
Problems 7.57
WOW! I am getting good!
Use the equation lambda=h/mv, where v is now speed because the question is asking the wavelength of a neutron.
6.63 x 10-34/1.67 x 10-27 (4.15) = 9.57 pm
Problem 7.58
6.63 x 10-34/1.67 x 10-27 (6.58) = 6.03 10-8, far ultra-violet
Part B. Matter-waves, the uncertainty principle and the Wave Model of the Atom
Problem 7.61 (pg. 290)
6.63 x 10-34/(145)(30.0) = 1.52 x 10-37 ( the answer is 1.52 x 10-22 --any idea what went wrong), decreased, smaller
The uncertainty value is since the electron is affected by the nucleus, in moves around inside a shell. For example, how earth goes around the sun. Even though the electron has an as exact orbit in an atom around the nucleus (or shell), the unknown of where how the electron will move is unknown value.
In figure 7.17 mostly you will find the hydrogen atoms electron in this region (pg.279). It is somewhere from 110 pm, and below from the lowest energy from only the wave.
In figure 7.18 the radius is shown of the probability of where the hydrogen atoms electron is from the nucleus. The peak in the figure is at 50 pm, which in the highest radial probability.




Part C. Wave Model Versus Bohr's Model
In think figure 7.10 is confusing. I do understand how the negative constant (-RH) is related to principal quantum number (n). As the n increase the -RH increases? Does anyone know why? The graph doesn’t show the spherical shape of the atom, instead it shows it in a straight line and maybe that’s why I do not like it.
In figure for 7.23 it is showing a 1s orbital. The electron distribution, or probability of finding the electrons is where the pink color is darker. In the 2s orbital the probability has two dark pink rings. Again, that denotes where the electron should be. In figure 7.24 shows the outer membrane (not sure what to call that) of the atom is least likely place you will find the electron. Since the diagram is a cutaway, you can see the darkest pink rings and that is where you will most likely find the electron.


If I was to redraw the Bohr I would use triangles. Each line would represent one n value. n=1 being the smallest triangle, n=2 second triangle, n=3 third triangle





























The Pauli Exclusion principle is that no two elements can have the same n number. For example each shell holds two electrons. One 1s orbital holds two electrons, one ms value is ½ the magnetic spin and the other ms value is – ½ the magnetic spin. The spin is represented in arrows. You can not have two up arrows in an orbit because of the electrons behavior to magnetic energy.

Exercise 8.1

A. Impossible, because 1s can hold 2 electrons.
B. Impossible, you have to fill up 1s first, then 2s second, the 2p
C. Impossible, two with the same spin
D. Possible, written correctly
E. Impossible, 2s can only fit 2 electrons
F. Impossible, after the 3p orbital is full, you have to put one electron in the 4s orbital to move on to the next orbital

Exercise 8.41

A. Impossible, two same spin
B. Possible, ok
C. Impossible, two same spin
D. Impossible, three electrons in one orbit (2 max)

Exercise 8.42
A. Impossible, two same spin
B. Possible, ok
C. Impossible, three electrons in one orbit
D. Possible,

Exercise 8.43
A. Impossible, 1s isn’t full
B. Impossible, 3s only room for two electrons
C. Possible, ok
D. Possible, d can hold 10


Hund's rule is that you must fill each subshell with a positive ms value. This is because they have low energy and have to be placed into different orbital subshells.
An example of an orbital diagram that obeys the rule is one page 309. [Ar] electron configuration (valance shell) is 1s22p63s63p6. If you use the Hund’s rule you can abbreviate [Ar] 3d64s2 . This is because according to the rule, you can fill up each orbital with one electron (in the same spin). Instead of pairing on the electrons in the orbital, they can be separated out.
The Aufbau principle is the building of the ground state of the electron configuration by filling the orbitals in its electron configuration. For example, 3s has lower energy than 3p. The longer the electron configuration the higher the energy it will be (build up principle). Another example is 2p2 has less energy than 2p6.
On pg 309 Neon obeys the rules, and C and N does not. (I think)
C & N= Paramagnetic Neon = Diamagnetic (definitely)
Other
An orbital that has two electrons is more stable than an orbital that has one electron.