Read Chapter 5
1.) Is it possible for a helium atom to have only 1 proton in its nucleus?
Answer: No. All helium atoms have 2 protons in their nucleuses (nuclei).
If an atom has only 1 proton it its nucleus, then it is a hydrogen atom.
(it behaves like other hydrogen atoms, not like a helium atom.)
2.) In order for an atom to have no net charge, it must have:
a.) the same number of protons as neutrons
b.) the same number of protons as electrons
c.) the same number of neutrons as electrons
Answer: b, the same number of protons as electrons.
3.) If you wanted to ionize a hydrogen atom, would you have to give
the atom energy, or take energy away from it?
Answer: You would have to give the atom energy.
4.)
For each statement below, write whether it is true or false:
a.) The water molecules in gas phase H2O have more kinetic energy
than the water molecules in solid phase H2O
Answer: True
b.) Most of the atoms in liquids are ionized
Answer: False
c.) The atoms in a plasma are neutral
Answer: False
d.) The temperature at which H2O
molecules dissociate is usually hotter than
the temperature of solid H2O
Answer: True
5.) Problem from the book:
See the energy level diagram for problem 42, in chapter 5 of the 4th
edition of the textbook (also at
http://hal.physast.uga.edu/~rls/1020/ch4/fig4-prob19.jpg)
a.) Which transition represents an electron that gains 10.2 eV of electrical
potential energy?
Answer: Transition B, which is from the 0.0 eV level to the 10.2 eV level
b.) Which transition represents an electron that loses 10.2 eV of electrical
potential energy?
Answer: Transition C, which is from the 10.2 eV level to the 0.0 eV level
c.) Which transition represents an electron that is breaking free of the
atom?
Answer: Transition E
d. Which transition is not possible?
Answer: "Transition" D, which does not go to a real energy level
e.) If an electron moved from level 3 to level 2, how much electrical
potential energy would the electron gain or lose (and, which is it:
gain energy or lose energy)?
Answer: The electron would lose (12.1 - 10.2) eV = 1.9 eV in energy
6.) The orange floor in the lecture hall could look orange to us because it
a.) emits orange light and no other visible colors of light
b.) absorbs orange light and no other visible colors of light
c.) reflects orange light and no other visible colors of light
d.) transmits orange light and no other visible colors of light
Answer: c
7.)
Problem: Compare visible light with radio waves:
a.) Which has the longer wavelength?
Answer: Radio waves are longer than visible light
b.) Which has the larger frequency?
Answer: Visible light has larger frequencies than radio waves
c.) Which has the larger energy (per "photon")?
Answer: The energy of a photon = h x frequency, so a visible photon has
more energy than a radio photon
8.)
Problem from textbook:
Summarize the circumstances under which objects produce
a.) thermal spectra
Answer: Thermal spectra are the same as "blackbody spectra".
They are produced by opaque (same as "optically thick") objects.
Light produced by atoms deep within the object is absorbed and re-emitted
by other atoms closer to the surface.
Several physical effects
(Doppler shifts, absorption and emission by a variety of
types of atoms or ionization states,
quantum mechanical effects),
change the wavelength of the photons, so that
the spectrum emitted by the object is a
continuous spectrum spanning a wide range of wavelengths.
b.) emission line spectra
Answer: Emission line spectra are emitted by optically thin
material. When an electron moves from a high energy level
to a low energy level, the atom emits light of a particular
wavelength. This light is called an emission line.
(For the purists
in the audience: in class and in the book we noted that a molecule
can also emit emission lines when the molecule shifts
from a higher rotational or higher vibrational level to a lower
level. Furthermore, there are other sorts of energy levels
and transitions between them also result in emission lines.)
The conglomerate of emission lines produced by an optically
thin material is called an emission line spectrum.
c.) absorption line spectra
Answer: If you shine a full (i.e. continuous) spectrum
of light on an optically thin material, then the optically
thin material
absorbs particular colors of light. The
absorbed photons have just the right energies to move
electrons within the atoms
from lower energy levels to higher energy levels
(or to make molecules rotate or vibrate at higher energy
levels, etc.).
The initial, continuous spectrum minus
the light absorbed by the optically thin material is
called an absorption line spectrum.
9.) Problem from textbook:
Suppose that the temperature of the Sun's surface were 12,000 K rather
than its current temperature (6,000 K).
a.) How much more thermal radiation would the Sun emit (i.e. what is
the ratio between the "new" quantity of emission
and the current quantity of emission)?
Answer: The ratio is 16. The reason is that power/area = sigma T4
So if we double T, then the Sun's power increases
by a factor of 24, which is 16.
b.) How would the color of the emission compare with the current color?
Answer: The Sun is optically thick, so its spectrum peaks at
wavelength = 2,900,000 nm / T (in Kelvins).
If T increases by a factor of 2, then
the spectrum will peak at a lower wavelength (by a factor of 2).
Compared with the
current spectrum, the new spectrum will be "bluer", though, technically,
the new spectrum peaks outside the visible part
of the spectrum.
10.) How fast does a spaceship with violet headlights
(the wavelength of this light is 400 nanometers)
have to travel in order for its light to look
orangy-red (the wavelength of this light is 600 nanometers)
to you?
a.) 1/6 c
b.) 1/4 c
c.) 1/2 c
d.) 2/3 c
e.) 3/2 c
Answer: c.
The Doppler shift formula applies here:
radial velocity / c = (observed wavelength - rest wavelength )/ rest wavelength
So,
radial velocity = c x [(observed wavelength - rest wavelength )/ rest wavelength]
So, radial velocity = c x [(600 nm - 400 nm) / 400 nm]
So, radial velocity = c x (200 nm / 400 nm)
radial velocity = (1/2) c
11.) Regarding the previous problem, which direction relative
to you would the
spacecraft have to be traveling in order for you to see this effect?
a.) Toward you
b.) Away from you
c.) Perpendicular to the line between you and it
d.) Any direction
Answer: b. The spacecraft is moving away from you.
Two lines of reasoning confirm this. Firstly, the
radial velocity calculation resulted in a positive radial velocity,
that means that the spacecraft is moving away from you (in contrast, a negative radial
velocity would have meant that the spacecraft was moving toward you).
Secondly, the wavelength appears to you to be longer than the
original wavelength. You can think of this as stretching the wavelength,
and stretching requires that the spacecraft is moving away from you
not toward you.
