Homework 2 Key

Q2.1

All of these parts are simple algebra calculations with the the de Broglie equation.

\[ \lambda = \dfrac{h}{p} = \dfrac{h}{mv} \]

a. A 100 kg man jogging on a treadmill with a velocity of 1 m/s

\[\lambda = \dfrac{h}{100 \;kg \times 1 \;m/s} = 6.626 \times 10^{-36} \;m \]

\[KE = .5(100 \;kg)(1 \;m/s)^{2} = 50 \;J\]

b. A 10 g silver bullet fired at a velocity of 1000 m/s

\[\lambda = \dfrac{h}{10 \;g \times 1000 \;m/s} = 6.626 \times 10^{-35} \;m\]

\[KE = .5(.01 \;kg)(1000 \;m/s)^{2} = 5000 \;J\]

c. The moon as it orbits the Earth

From Wikipedia \(m_{moon} = 7.3 \times 10^{22} \;kg \) and \(v_{orbital} = 1022 \;m/s \)

\[\lambda = \dfrac{h}{7.3 \times 10^{22} \;kg \times 1022 \;m/s} = 8.881 \times 10^{-60} \;m\]

\[KE = .5(7.3 \times 10^{22} \;kg)(1022 \;m/s)^{2} = 3.8 \times 10^{28} \;J\]

d. A positron with a velocity of \(7 \times 10^6\) cm/s

A positron is an electron with a positive charge (Simplification - don't tell a physicist)

\[\lambda = \dfrac{h}{9.1 \times 10^{-31} \;kg \times 7 \times 10^4 \;m/s} = 1.04 \times 10^{-8} \;m = 10.4 \;nm\]

\[KE = .5(9.1 \times 10^{-31} \;kg)(7 \times 10^{4} \;m/s)^{2} = 2.2 \times 10^{-21} \;J\]

The first result that has a meaningful wavelength.

e. A chicken crossing the road with a velocity of 1 cm/s. Assume the chicken has a mass of 1 kg.

\[\lambda = \dfrac{h}{1 \;kg \times 0.01 \;m/s} = 6.626 \times 10^{-32} \;m\]

\[KE = .5(1 \;kg)(.01 \;m,s)^{2} = 5 \times 10^{-5} \;J\]

f. An electron in the n = 1 Bohr orbital which has a velocity of \( 2.19 \times 10^6\) m/s

The \(n=1\) Bohr orbital is extra information we don't need.

\[\lambda = \dfrac{h}{9.1 \times 10^{-31} \;kg \times 2.19 \times 10^6 \;m/s} =3.32 \times 10^{-10} \;m = 0.332 nm \]

\[KE = .5(9.1 \times 10^{-31} \;kg)(2.19 \times 10^{6} \;m/s)^{2} = 4.2 \times 10^{-18} \;J\]

In all these deBroglie wavelenght questions pay attention to when the wavelength actually matters. Only when the object is very small with a mass of about the same order \( 10^{-30}\) ish as \(h\) is the wavelength important. Quantum effects became too small to ever notice with everyday "real" objects.

Q2.2

Estimate the de Broglie wavelength of electrons that have been accelerated from rest through a potential difference (V) of 100 kV. (Hint: kinetic energy is eV).

The de Broglie Wavelength is found by:

\[ \lambda = \dfrac{h}{mv} = \dfrac{h}{p} \]

We know the kinetic energy is given by:

\[E_K = eV = \dfrac{1}{2} mv^2 = \dfrac{p^2}{2m} \]

And can be related to the momentum:

\[p= \sqrt{2mE_K} \]

Therefore:

\[ \lambda = \dfrac{h}{\sqrt{2meV}} \]

Input the values:

\[ \lambda = \dfrac{6.63 \times 10^{-34}\; J \cdot s}{\sqrt{2(9.1 \times 10^{-31} \; kg)(1.6 \times 10^{-19} \; C)(100 \times 10^{3} \;V)}} \]

And the wavelength is:

\[ \lambda = 3.8 pm \]

Q2.3

A thermalized electron is an electron with the kinetic energy given by \(k_BT\) with \(k_B\) as the Boltzmann constant. Calculate the de Broglie wavelength of a thermalized electron at 295 K.

This problem is the same as Number 2, except the kinetic energy is now:

\[E_K = k_BT \]

Therefore from equation (Q1 4) we have:

\[ \lambda = \dfrac{h}{\sqrt{2mk_BT}} \]

Input the values:

\[ \lambda = \dfrac{6.63 \times 10^{-34}\; J \cdot s}{\sqrt{2(9.1 \times 10^{-31} \; kg)(1.38 \times 10^{-23} \; J/K)(295 \;K)}} \]

And the wavelength is:

\[ \lambda = 7.7 nm \]

Q2.4

Estimate the theoretical resolution of an electron microscope that produces electrons from a tungsten filament with an energy of 200 keV.

Hint: The resolution of a microscope is the shortest distance between two points on a sample that can still be distinguished by the observer or camera system as separate entities. This is best estimated as half the wavelength of the incident particles for this discussion although more complex formulations exist.

This question is basically the same as Number 2 with a different energy value.

\[ \lambda = \dfrac{h}{\sqrt{2meV}} \]

Input the values:

\[ \lambda = \dfrac{6.63 \times 10^{-34}\; J \cdot s}{\sqrt{2(9.1 \times 10^{-31} \; kg)(1.6 \times 10^{-19} \; C)(200 \times 10^{3} \;V)}} \]

And the wavelength is:

\[ \lambda = 2.7 pm \]

And the resolution is half the wavlength

\[\; Resolution = \dfrac {2.7 \;pm }{2} = 1.35 \; pm\]

Q2.5

The Pashen series is:

\[ \bar{\nu}= \dfrac{1}{ \lambda} =R\left( \dfrac{1}{9} -\dfrac{1}{n_2^2}\right), n_2>3 \]

And the highest energy line will occur when the electron relaxes from the highest possible energy state. Thus:

\[n_2 = \infty \]

Input the values:

\[ \dfrac{1}{ \lambda} =(1.09 \times 10^{7}\; 1/m)\left( \dfrac{1}{9} -\dfrac{1}{\infty}\right) \]

The wavelength is:

\[ \dfrac{1}{ \lambda} = 1211111 \; 1/m \]

\[\lambda = 825 nm \]

And the lowest energy line will occur when the electron relaxes from the lowest possible energy state. Thus:

\[n_2 = 4 \]

Input the values:

\[ \dfrac{1}{ \lambda} =(1.09 \times 10^{7}\; 1/m)\left( \dfrac{1}{9} -\dfrac{1}{16}\right) \]

The wavelength is:

\[ \dfrac{1}{ \lambda} = 529861 \; 1/m \]

\[\lambda = 1887 nm \]

In the Bohr atom, angular momentum is quantized

\[L = m_evr = n\hbar \]

And the lowest energy state is

\[n=1 \]

so:

\[L = \hbar \]

which means that the velocity is:

\[v = \dfrac{\hbar}{m_er} \]

Input the values:

\[v = \dfrac{1.05 \times 10^{-34} \; J \cdot s}{(9.1 \times 10^{-31} \; kg)(5.29 \times 10^{-11} \; m)} \]

And the velocity is:

\[v = 2.2 \times 10^6 \; m/s \]

Q2.6

The Heisenberg Uncertainty Principle says:

\[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi} \]

If we insert the definition for momentum:

\[m\Delta{v}\Delta{x} \ge \dfrac{h}{4\pi} \]

\[\Delta{v} \ge \dfrac{h}{4m\pi\Delta{x}}\]

Now input the values for the arrow problem:

\[\Delta{v} \ge \dfrac{6.63 \times 10^{-34}\; J \cdot s}{4(0.5 \; kg)\pi(1.0 \times 10^8\;m)} \]

And

\[\Delta{v} \ge 1.05 \times 10^{-42} \; m/s \]

Note: I think the problem is meant to have \(\Delta{x} = 1 \times 10^{-8} \; m \) which would be more sensible for a baseball pitch. Then the answer is \(\Delta{v} \ge 1.05 \times 10^{-26} \; m/s \)

The Heisenberg Uncertainty Principle only limits the combined minumim uncertainty to be greater than or equal to \( \dfrac{h}{4\pi} \) there is no limit on the maximum. The maximum can be as big as you want. No math required.

\[\Delta{x} = \infty\]

Q2.7

A quantum mechanical turtle crosses the road at a velocity of \(1 \times 10^{16}\) nm/sec, with an uncertainty of \(10^{-283}\) mm/sec. Calculate the uncertainty in location of the turtle. (Assume the turtle has a mass of 1 kg.)

The Heisenberg Uncertainty Principle says:

\[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi} \]

If we insert the definition for momentum:

\[m\Delta{v}\Delta{x} \ge \dfrac{h}{4\pi} \]

Rearrange for \(\Delta{x}\)

\[\Delta{x} \ge \dfrac{h}{4m\pi\Delta{v}} \]

Input the values in the correct units m/sec and kg for the turtle.

\[\Delta{x} \ge \dfrac{h}{4 (1 \; kg) \pi (1.0 \times 10^{-286} \;m/s)} \]

\[\Delta{x} \ge \pi\times 10^{252} \;m \]

(Note the size of the observable universe is only 10^26 m. So the turtle could be anywhere, literally.)

Q2.8

To show that a specific expression is a solution to a differential we have to plug it in and see if it works.

The function is:

\[U(x,t) = e^{i(k x + ωt)} \]

And the classical wavefunction is:

\[\dfrac{\delta^{2}U}{\delta x^2} = \dfrac{1}{v^2} \dfrac{\delta^{2}U}{\delta t^2} \]

\[k^2 e^{i(k x + ωt)} = -\dfrac{\omega^2}{v^2} e^{i(k x + ωt)} \]

So \(e^{i(k x + ωt)}\) is a solution subject to \(–k^2 = \dfrac{\omega^2}{v^2}\)

(ii) The function is now:

\[U(x,t) = \cos(k\,x - \omega\, t) \]

\[\dfrac{\delta^{2}U}{\delta x^2} = \dfrac{1}{v^2} \dfrac{\delta^{2}U}{\delta t^2}\]

\[-k^2 \cos(k\,x - \omega\, t) = \dfrac{\omega^2}{v^2} \cos(k\,x - \omega\, t)\]

So \(\cos(k\,x - \omega\, t)\) is a solution subject to \(–k^2 = \dfrac{\omega^2}{v^2}\)

Q2.9

Part I: So first we set 2.1 equal to 2.2 and use the trig identity given to expand out 2.2

\[A \cos(\omega t) + B \sin(\omega t) = C \cos (\omega t + \Phi) = C \cos (\omega t) \cos (\Phi) - C \sin (\omega t) \sin (\Phi) \]

Because A, C, B, and Φ are constants. This equation works as long as \(C \cos(\Phi) = A\) and \(C \sin(\Phi) = B \) which is the answer to part III.

Again for 2.3 we do the same thing

\[A\cos(\omega t) + B\sin(\omega t) = C\sin(\omega t + \Psi) = C\sin(\omega t)\cos(\Psi) – C\cos(\omega t)sin(\Psi) \]

Because A, C, B, and Ψ are constants. This equation works as long as \(C\cos(\Psi) = B \) and \(C\sin(\Psi) = A\) which is the answer to part IIII.

Part II: To show that 2.2 and 2.3 are solutions we simply plug them into the differential equation.

(2.2)

\[\dfrac{\delta U^2}{\delta t^2} + \omega^2 U = 0\]

\[U(t) = C\cos(\omega t + \Phi) \]

\[-\omega^2 C \cos(\omega t + \Phi) + \omega^2 C \cos (\omega t + \Phi) = 0 \]

(2.3)

\[\dfrac{\delta U^2}{\delta t^2} + \omega^2 U = 0\]

\[U(t) = C\sin(\omega t + \Psi) \]

\[-\omega^2 C \sin(\omega t + \Psi) + \omega^2 C \sin (\omega t + \Psi) = 0 \]

Q2.10

Wave equation between 0 and L

\[ \dfrac{\partial^2 u(x)}{\partial x^2} + \left( \dfrac{8\pi^2m E}{h^2} \right) u(x)= 0 \]

with these boundary conditions:

\[u(0)= u(L) = 0 \]

This is a particle in a box. Use the Dead French Mathematician method.

\[u(x) = A \sin kx + B \cos kx\]

First Boundary condition \(u(0) = 0\) means that B must equal zero.

Second Boundary condition \(u(L) = 0\) means that \(kL = nπ\) since \(A \sin(kL) = 0\) thus

\[k = \dfrac{nπ}{L}\]

Additionally from solving the differential equation we get \(k^2 = \dfrac{8π^2mE}{h^2}\) we can combine the two equations to eliminate k.

\[ \dfrac{8π^2mE}{h^2} = \dfrac{n^2π^2}{L^2}\]

Then solving for \(E\) to get

\[E = \dfrac{n^2h^2}{8mL^2} \]

The energy is quantized because E can only take certain values given by n = 1, 2, 3, 4, …

Q2.11

The total energy is the sum of the Kinetic and Potential energies (E = T + V). The exponential form of the solution to the wave equation is:

\[ \Psi = a e^{2\pi i x / \lambda} e^{-2\pi i \nu t} \]

And the de Broglie relation is:

\[\lambda = \dfrac{h}{mv} = \dfrac{h}{p} \]

To rewrite the generalized 1D time-dependent wave equation in terms of total energy we need to replace the wavelength using de Broglie.

\[T = \dfrac{p^2}{2m} \]

\[p = \sqrt{2mT} \]

\[\lambda = \dfrac{h}{\sqrt{2mT}}\]

\[T=E-V\]

Finally

\[\lambda = \dfrac{h}{\sqrt{2m(E-V)}} \]

The wavefunction is written

\[ \Psi = a e^{\dfrac{i 2\pi xh}{\sqrt{2m(E-V)}}} e^{-2\pi i \nu t} \]

So now lets examine what happens in 2 cases

If \(E > V\) then \(\sqrt{2m(E-V)}\) is positive and

\[ \Psi = a e^{i K x} e^{-2\pi i \nu t} \]

Which is an oscillating function over time. It looks like sine or cosine wave.

If \(E < V\) then \(\sqrt{2m(E-V)}\) is negative and

\[ \Psi = a e^{i^2 K x} e^{-2\pi i \nu t} \]

Where \(i^2 = -1\) then we have

\[ \Psi = a e^{-K x} e^{-2\pi i \nu t} \]

which is a decaying exponential over time.