ANSWER: The longest wavelength (ground) state of
a particle in a one-dimensional (1D) "box"
has and therefore (by de Broglie's hypothesis)
momentum
. For a "1D box" this large
the electron will be nonrelativistic (you can check this),
so we can set
and thus
where
is the time
for the electron to make a round trip to the other end
and back. Thus the electron transfers momentum
(by reversing its direction) to one end of the "string"
every
, generating an average force
.
Then we just plug in
m,
kg and
J-s to get
or
.
Since each end of the SWNT is being pushed "out"
by this force, it is the same as the tension
in the SWNT.1
ANSWER: The arguments developed for the time-averaged force
exerted on one wall of a cubical 3D container apply equally well
for the force acting on the boundary of a 1D "box" - namely,
for a single particle bouncing back and forth,
,
where
is the length of the "box".
In this case there is only one direction of motion
(degree of freedom) so we can drop the
subscript on
.
The EQUIPARTITION THEOREM says that the thermal
average of the kinetic energy
associated with
this translational degree of freedom is
, giving
.
If we substitute this back into the formula for
we get
. For
particles all
doing this at the same time, we just multiply by
, giving
or (ignoring the fact that the
actual force
at any instant fluctuates minutely about the average
force
)
.
This is the 1D equivalent of the IDEAL GAS LAW.
Note that the only difference between this derivation and the one for
a 3D box is that here we didn't have to introduce the notion of
pressure as the force per unit area. This is simpler!
ANSWER: Again the 1D case is much simpler than the 3D case
because we don't have to worry about
composing out of
(or
out of
,
and
).
There is just one direction, just one velocity component,
one momentum component and one wavelength to worry about
making commensurate with the length of the "box".
Thus the requirement that an integer
half-wavelengths
fit evenly into
gives
and so
or
. Thus the possible values of
the speed
are evenly spaced every
and the distribution of speeds varies only as the
Boltzmann factor
. This gives immediately
where
is a normalization constant that does not depend on
.
You get full credit for this result, but here's how to get
:
Let
where
is treated as a constant.
Now,
or
so
and we have
You can look up the definite integral; its value is
,
giving
.
ANSWER:
As stated in the textbook, the most probable speed for an
ideal gas molecule in 3D is
(i.e. when
).
For the 1D gas, however,
is missing that
extra factor of
that forces its value to zero at
,
so the resultant speed distribution has its maximum
at
:
.