1. a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. b) If a photon and an electron each have the same wavelength of 250 nm, find the energy of each.
c) You want to study an organic molecule that is about 250 nm long using either a photon or an electron microscope. Approximately what wavelength should you use and which probe, the electron or the photon, is likely to damage the molecule least?
2. Electrons in a beam incident on a crystal at an angle of 300 have kinetic energies ranging from zero to a maximum of 5500 eV. The crystal has a grating space d=0.5A0, and the reflected electrons are passed through a slit .Find the velocities of the electrons passing through the slit. How many are these velocities?
3. a) Show that the phase velocity of the de Broglie waves of a particle of mass m and de Broglie wavelength is given by .
b) Compare the phase and group velocities of an electron whose de Broglie wavelength is exactly 1x10-13 m.
4. To what voltages must we accelerate electrons (as in an electron microscope, for example) if we wish to resolve;
a) a virus of diameter 12 nm , b) an atom of 0.12 nm, c) a proton of diameter 1.2 fm?
5. a) The x-coordinate of an electron is measured with an uncertainty of 0.20 mm. What is the x- component of the electron’s velocity, vx, if the minimum percentage uncertainty in a simultaneous measurement of vx is 1.0%.
b) Repeat part (a) for a proton. 6. Suppose Fuzzy, a quantum mechanical duck, lives in a world in which h = 2 J.s. Fuzzy has a mass
of 2.0 kg and is initially known to be within a region 1.0 m wide.
a) What is the minimum uncertainty in his speed? b) Assuming this uncertainty in speed to prevail for 5.0 s, determine the uncertainty in position after
this time.
7. The seeing ability, or resolution, of radiation is determined by its wavelength. If size of an atom is of the order of 0.1 nm, how fast must an electron travel to have a wavelength small enough to “see” an atom?
8. An air rifle is used to shoot 1.0 g particles at 100 m/s through a hole of diameter 2.0 mm. How far from the rifle must an observer be to see the beam spread by 1.0 cm because of the uncertainty principle? Compare this answer with the diameter of the universe ( 1026 m).
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1. a) To find the wavelength of a photon or an electron with a given energy, we can use the energy-wavelength relationship:
For photons:
The energy of a photon is given by the equation E = hc/λ, where E is the energy, h is the Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength. Rearranging the equation, we have λ = hc/E.
Substituting the given values (energy = 20.0 eV = 20.0 x 1.602 x 10^-19 J) into the equation, we have:
λ = (6.626 x 10^-34 J.s x 3.00 x 10^8 m/s) / (20.0 x 1.602 x 10^-19 J)
Simplifying the equation will give you the wavelength of the photon.
For electrons:
The energy of an electron is given by the equation E = (1/2)mv^2, where E is the energy, m is the mass of the electron (9.11 x 10^-31 kg), and v is the velocity. We can use the de Broglie wavelength equation λ = h / (mv).
Substituting the given values (energy = 20.0 eV = 20.0 x 1.602 x 10^-19 J) into the equation, we have:
λ = (6.626 x 10^-34 J.s) / ((9.11 x 10^-31 kg) x v)
Further simplification will give you the wavelength of the electron.
b) To find the energy of a photon or an electron with a given wavelength, we can use the energy-wavelength relationship mentioned earlier.
For photons:
Rearranging the equation E = hc/λ, we have E = hc/λ.
Substituting the given values (wavelength = 250 nm = 250 x 10^-9 m) into the equation, we have:
E = (6.626 x 10^-34 J.s x 3.00 x 10^8 m/s) / (250 x 10^-9 m)
Simplifying the equation will give you the energy of the photon.
For electrons:
Rearranging the equation λ = h / (mv), we have v = h / (mλ).
Substituting the given values (wavelength = 250 nm = 250 x 10^-9 m) into the equation, we have:
v = (6.626 x 10^-34 J.s) / ((9.11 x 10^-31 kg) x (250 x 10^-9 m))
Further simplification will give you the velocity of the electron. Since kinetic energy is given as (1/2)mv^2, you can calculate the energy of the electron.
c) To study an organic molecule of length 250 nm, you would want a wavelength of the same scale to effectively observe the molecule. Both photon and electron microscopes can achieve this.
Photon microscopy uses light waves and can achieve wavelengths in this range by using visible light or ultraviolet light. The advantage of using photons is that they have lower energy compared to electrons. This makes them less likely to damage the organic molecule as they interact with it.
Electron microscopy, on the other hand, uses accelerated electrons to form an image. While electrons can achieve smaller wavelengths than photons, they have higher energy and can potentially damage the organic molecule, especially if the acceleration voltage is high. Therefore, in terms of damaging the molecule least, using a photon microscope would be a safer choice.