Thursday, February 24, 2011

Mastering Physics: ± Accelerating Electrons

Part A
Through what potential difference DeltaV must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.160 nm?
Use 6.63×10−34 J \cdot s for Planck's constant, 9.11×10−31 kg for the mass of an electron, and 1.60×10−19 C for the charge on an electron.
ANSWER:

  DeltaV  = 58.8
  \rm V
 
Part B
Through what potential difference deltaV must electrons be accelerated so they will have the same energy as the x-ray in Part A?
Use 6.63×10−34 J \cdot s for Planck's constant, 3.00×108 m/s for the speed of light in a vacuum, and 1.60×10−19 C for the charge on an electron.
ANSWER:

  DeltaV  = 7760
  \rm V

Mastering Physics: ± Uncertainty in the Atomic Nucleus

Rutherford's scattering experiments gave the first indications that an atom consists of a small, dense, positively charged nucleus surrounded by negatively charged electrons. His experiments also allowed for a rough determination of the size of the nucleus. In this problem, you will use the uncertainty principle to get a rough idea of the kinetic energy of a particle inside the nucleus.
Consider a nucleus with a diameter of roughly 5.0 \times 10^{-15} meters.


Part A
Consider a particle inside the nucleus. The uncertainty Deltax in its position is equal to the diameter of the nucleus. What is the uncertainty Deltap of its momentum? To find this, use \Delta x \Delta p \ge \hbar.
Express your answer in kilogram-meters per second to two significant figures.
ANSWER:

  Deltap  = 2.10×10−20
  \rm kg \cdot m/s

Part B
The uncertainty Deltap sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum were to fall below that point, then the uncertainty principle would be violated. Since the uncertainty principle is a fundamental law of physics, this cannot happen. Using \Delta p = 2.1 \times 10^{-20} kilogram-meters per second as the minimum momentum of a particle in the nucleus, find the minimum kinetic energy K_min of the particle. Use m=1.7 \times
10^{-27} kilograms as the mass of the particle. Note that since our calculations are so rough, this serves as the mass of a neutron or a proton.
Express your answer in millions of electron volts to two significant figures.
ANSWER:

  K_min  = 0.81
  \rm MeV
Compare this to the normal energy scale for electrons in an atom, which is on the order of single electron volts. The characteristic energy scale for the nucleus seems to be roughly one million times that for electrons in an atom. This difference can be seen in calculations of the energy output per unit mass from coal-burning power plants, which utilize chemical energy (energy associated with the electrons of an atom) compared to the energy output from nuclear reactors (power plants that harness the much higher energies of nuclei).

Mastering Physics: ± Simultaneous Measurements of Position and Velocity

Part A
The x coordinate of an electron is measured with an uncertainty of 0.200 mm. What is v_x, the x component of the electron's velocity, if the minimum percentage uncertainty in a simultaneous measurement of v_x is 1.00 \%? Use the following expression for the uncertainty principle:
\Delta x \Delta p_x \ge \hbar,
where \Delta x is the uncertainty in the x coordinate of a particle, \Delta p_x is the particle's uncertainty in the x component of momentum, and \hbar = \frac{h}{2\pi}, where h is Planck's constant.
Express your answer in meters per second to three significant figures.
ANSWER:

  v_x  = 57.9
  \rm m/s


Part B
Repeat Part A for a proton.
Express your answer in meters per second to three significant figures.
ANSWER:

  v_x  = 3.16×10−2
  \rm m/s

Mastering Physics: ± Resolution of an Electron Microscope

An electron microscope is using a 1.00-keV electron beam. An atom has a diameter of about 10^{-10} meters.

Part A
What is the wavelength lambda of electrons in this microscope?
Express your answer in nanometers to three significant figures.
ANSWER:

  lambda  = 3.88×10−2
  {\rm nm}
Compare this wavelength to the typical size of an atom.
 

Part B
Can an individual atom theoretically be resolved using this electron microscope?

ANSWER:



Note that in practice a resolution better than \sim 0.5 \:{\rm nm} cannot be achieved with an electron microscope. This is due, in part, to the fact that the focal length of a magnetic lens depends on the electron speed, which is never the same for all electrons in the beam.
 
 
Part C
Suppose we replace the electron beam with a proton beam. What proton energy is needed to achieve the same resolution as the electron beam in Part A?
Express your answer in electron volts to three significant digits.
ANSWER:


0.545
  {\rm eV}
In principle, we could use proton beams with relatively low energy to get great resolution. However, in practice, it is much easier to create an electron beam than it is to create a proton beam, which is why electron microscopes have become the norm.
 

Mastering Physics: Width of a Wave Function

A particle is described by a wave function  \psi (x)=Ae^{- \alpha x^{2}}, where A and alpha are real, positive constants.

Part A
If the value of alpha is increased, what effect does this have on the particle's uncertainty in position?
ANSWER:


Part B
If the value of alpha is increased, what effect does this have on the particle's uncertainty in momentum?

 ANSWER: