Physics 150 Fall 2003

Solution to HW 8

Grading: 27 points total: 2 points for each problem except 4 points for #1 (2 points for parts A and C. Part

B was not graded since it was not covered in class) and 3 points for #2.

1. Question A,

(1) When heat is added to a liquid the atoms move at larger speeds. When enough heat is added to raise the temperature to the boiling point the atoms have enough energy to break free and form a gas. In the gas the atoms are well separated and move freely except when they collide.

(2) The liquid is formed because the atoms are bound together with each atom surrounded by other atoms. At the surface the atoms are less bound because they have fewer other atoms around them. Thus the liquid tries to adopt a shape with the least possible surface area. This is called “surface tension” when the atoms try to come together to minimize the surface area.

(3) The atoms are pushed together so that they collide more often causing them to move more rapidly thus rasing the temperature.

2. There are many answers. Possible ones:

The way that atoms explain the formation of solids, liquids and gasses in terms of motion of the atoms.

The periodic table that shows all matter is described by
only a fixed set of types of elements.

The law of constant proportions when compounds are formed.

Scattering experiments that show particle being deflected by nuclei in a solid: the nuclei are extremely small but contain most of the mass.

Scanning tunneling microscope pictures that show individual atoms.

3. Ex. 1, Ch. 14 of
March: If the gold atoms have diameter
of 0.25 x 10^{-9} m, then in a a thicknesss of 10^{-6} m, there are 10^{-6}/0.25
x 10^{-9} = 4000 atoms.

4. Ex. 2, Ch. 14 of
March: Each alpha particle passes through
4000 atoms (previous problem). Each atoms can be
considered a target with the chance of hitting to nucleus equal to the area of
the nucleus divided by the area of the atom (because a target is an area). For
each atom the ratio is (2.0 x 10^{-14}/0.125 x 10^{-9})^{ 2}
= (16.0 x 10^{-5})^{ 2} = 256 x 10^{-10 }for each
atom. Thus the chance of hitting one
atom is 256 x 10^{-10 }x 4000 = 1024 x 10^{-7 }= 1.024 x 10^{-4 }= 0.01024 %

5. Ex. 4, Ch. 14 of March:
Using the formula the frequency = nu = 3.3 x
10^{15}(1/1^{2} -1/2^{2}) = 3.3 x 10^{15}(3/4) = 2.475 x 10^{15} Hz

6. Ex. 1, Ch. 15 of March: Using h = 4.15 x 10^{-15}
eV s, a photon of frequency 3.3 x 10^{15}
Hz has energy 4.145 eV.

7. Ex. 1, Ch. 15 of March: One way to work the problem is
change wavelength lambda
to frequency = c/ lambda and the use the same relation as in the
previous problem. 0.4 mm corresponds to frequency nu = 3 x
10^{8}/0.4 x 10^{-6} = 7.5 x 10^{14} Hz,
and 0.7 mm corresponds to nu = 3 x 10^{8}/0.7
x 10^{-6} = 4.3 x 10^{14} Hz.
These are photons with energy E = h nu = 4.15 x 10^{-15}
eV s x 7.5 x 10^{14} Hz = 3.11 eV and 4.15 x
10^{-15} eV s x 4.3 x 10^{14} Hz
= 1.77 eV
respectively.

8. Ex. 4 8, Ch. 15 of March: We can use either the Balmer formula or the formula on page 185. Since the frequency is asked for, it is easier to use the Balmer formula. This gives:

a.
(Same as Ex. 4, ^{15}(1/1^{2} -1/2^{2})
= 2.475 x 10^{15} Hz

b. nu = 3.3 x 10^{15}(1/2^{2} -1/3^{2})
= 4.58 x 10^{14} Hz

9. Question B, Ch. 16
of March: The basic formula is lambda = h/p.
One way to see the result is to first consider ordinary non-relativistic
particles of different mass with the relation p = mv
and E = (1/2) mv^{2} = (1/2) p^{2} /m. Thus for the same kinetic energy a light mass
particle has larger wavelength. The
reasoning can be applied to light by noting that is corresponds to a particle
of very light mass. One can also derive the wavelength using the formulas for
light (E=pc or p = E/c) and for an electron (E = {1/2} p v or p = 2 E/v)).

Thus for light lambda = hc/E and for an electron lambda = (1/2) hv/E. If v is much less than c, the wavelength for light is much longer than for an electron of the same energy.

10. Ex. 1,

This is just using de Broglie’s arguments in reverse. Using p=h/lambda, p=E/c and E=h nu, we find h nu/c= h/lambda or c = lambda nu, which is the correct formula for a wave.

11. Ex. 3, Ch. 16 of March: Using lambda = h/mv, we find lambda = 6.6 x 10^{-34} J s / 9 x

10^{-31} Kg x 10^{7} m/s = (6.6/9) x 10^{-10}
m = 7.3 x 10^{-9} m = 7.3 nm (roughly the size of an atom)

12. Ex. 4, Ch. 16 of March: lambda = 6.6 x 10^{-34}
J s / 6.6 x 10^{-27} Kg x 0.09 x10^{8} m/s = (1/0.09) x
10^{-15} m = 1.1 x 10^{-14} m (roughly the size of a nucleus)

13. EXTRA. Ex. 7, Ch.
15: All energies are proportional to the
mass (see formula on p. 185). Thus the
energies in the muonium atom (proton and muon) is 207 time that off the hydrogen atom, and 13.6 eV is replaced by 13.6 x 207 = 2815 eV. The ground state energy is – 2815 eV. The energy of
.transition from the state with n=2 to the lowest state with n=1 is Delta E =
2815 (1/1^{2} -1/2^{2}) = 2111 eV
which is a frequency of 4.15 x 10^{15} x 2111 = 8.76 x 10^{18}
Hz or a wavelength of lambda = c/nu = 3.42 x 10^{-9 }m.