UNIT 5 – QUANTUM MECHANICS and SPECIAL RELATIVITY
Homework
Chapter 12 Review
Q: 1-20, 23, 24 pg 659, 660
1. F – The wavelength of a single quantum is directly proportional to its frequency
2. T
3. F – For a given photoelectric surface, the shorter the wavelength, the higher the cut off potential
4. T
ℎ
5. F – Phtons have momentum whose value is given by p = 𝜆
6. F – When light passes through a medium, its behaviour is best explained using its particle properties,
whereas when light interacts with matter, its behaviour is best explained using its particles
characteristics.
7. T
8. F – Electrically excited gass produce an emission spectrum or line spectrum, while a continuous
spectrum is produced from aheated solid
9. T
10. T
11. F – In an atom we think of the electron as a particle moving in a circular orbit, whose wave
properties predict the probability of its position and velocity
12. d
Increasing the frequency of the light incident on a metal surface:
- Has no effect on the threshold frequency for the emission of photoelectrons
- Increases the number of photoelectrons emitted
- Decreases the threshold frequency for the emission of photoelectrons
- Increases the kinetic energy of the most energetic photoelectrons
- Increases the kinetic energy of the photoelectrons
13. b
The light beam that produced graph B had the highest frequency becuase it required the largest
negative potential difference to reduce its photoelectric current to zero
14. c
The longest wavelength is associated with the lowest frequency. Thus, the beam having the lowest
frequency must have produced curve C, becuase it fell to zero current with the lowest ngative potential
difference
15. b
Photons with highest momentum have the highest kinetic energy. These high energy electrons would
have require the highest negative potential difference to be stopped. Thus the ligt beam of graph B
must have produced them.
16. a
The formation of these rings is similar to that produced by waves passing through a small opening and
would be similar to that produced by light passing through a grating, if the grating were rotated through
360 degrees. Thus, the electrons diffraction pattern is similar to that for light, illustrating the wave
nature of electrons scattered by a thin gold film.
17. e
Since p2 = 2mEk, increasing the energy increases the momentum of the electrons. Since λ = h/p,
increasing p decreases wavelength. For diffraction grating λ/d = sin θ, decreasing λ decreases θ, and
thus decreasing the distance of the rings from the axis of the beam
18. d
As the 8 eV electron collides with the atom, energy ca be absorbed as follows:
8 – 6.67 = 1.33 eV
8- 4.86 = 3.12 eV
The electron could bounce off in an elastic collision, emerging with 8 eV
19. d
Since the 9 eV electron has energy exceding that of all of the energy levels shown in the diagram,
energies of 4.6 eV, 6.67 eV and 8.84 eV are all possible
20. e
For a photon to be absorbed, it must have exactly the sam energy, as energy levels in the atom. It does
not, so the atom can not be excited by a 9 eV photon
23. a
For n =3, E3 = 12 eV
n =2, E2 = 8 eV
Eemitted photon = E3 – E2
= 12 – 8
= 4 eV
24. b
The electron can transfer enough energy to the atom to raise the level from one state to another state,
with a higher energy level. In this case, the only possible transistion is from n=1 (0 eV) to n =2 (8 eV).
Thus, the electron transfers 8 ev to the atom and rebounds with 2 eV remaining
Q: 1, 4-8, 10, 14, 19, 26, 27 pg 661
1. Visible photons are the ones our eyes can detect. A dark room is full of photons with infrared
wavelength and greater, nut nore have sufficient energy to activate the nerve cells in the retina of our
eyes. Therefore, we cannot see in the dark because visible light photons do not exist. There may also
be photons of ultraviolet wavelengths and shorter, but these are also invisible to our eyes.
14. To analyze the composition of a gas, an absorption spectrum is used. Since the light arriving from
Mars to Vnus must pass through Earth’s atmosphere, no conclusive reults are possible since Earth’s
atmosphere is 20 % oxygen.
19.
Water Waves
Low speed
Large slits
Interference patterns spread over large surface
Diffraction is easy to utilize
Electrons
High speed
Very small slits (spaces between atoms)
Interference pattern very small
Diffraction is difficult, very short wavelengths
26. Applying the principle of uncertainty to an electron orbiting a nucleus we say that there is
uncertainty in measuring the electron’s position and velocity. Thus it becomes impossible to say for any
individual electron where it is now or where it will be at any future time. Rather than saying where it is,
we must be satisfied with describing it slocation by starting the probability that iw will be found near
any point. One way of visualizing this situation is to think of the electron existing as a cloud of negtive
charge distributed around the atom, rather than as a particle moving in a circular orbit. The cloud is
denser in areas of high probability and less dense in areas where the electron is less likely to be found.
27. An object can never be truly at rest. The equation ∆x∆p ≥ h/2π indicates that there is uncertainty in
any measurement of position, speed, and momentum. If the position is uncertain, then the object can
never be at rest, at least as we know it in classical mechanics. In reality, for macroscopic objects the
uncertainty is so small that we do not notice it in normal everyday life.