#2 Solutions

Q1

Estimate the de Broglie wavelength of electrons that have been accelerated from rest through a potential difference (V) of 30 kV. (Hint: kinetic energy is eV).

Solution

The de Broglie Wavelength is found by:

\[ \lambda = \dfrac{h}{mv} = \dfrac{h}{p} \tag{1} \]

We know the kinetic energy is given by:

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

And can be related to the momentum:

\[p= \sqrt{2mE_K} \tag{3}\]

Therefore:

\[ \lambda = \dfrac{h}{\sqrt{2meV}} \tag{4}\]

Input the values:

\[ \lambda = \dfrac{6.63 \times 10^{-34}\; J \cdot s}{\sqrt{2(9.1 \times 10^{-31} \; kg)(1.6 \times 10^{-19} \; C)(30 \times 10^{3} \;V)}} \tag{5}\]

And the wavelength is:

\[ \lambda = 7.09 pm \tag{6}\]

Q2

Calculate the de Broglie wavelength of a room temperature thermalized neutron (i.e., having a kinetic energy of \(k_BT\) with \(k_B\) as the Boltzmann constant).

Solution

This problem is the same as Number 1, except the kinetic energy is now:

\[E_K = k_BT \tag{1}\]

Therefore from equation (Q1 4) we have:

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

Input the values:

\[ \lambda = \dfrac{6.63 \times 10^{-34}\; J \cdot s}{\sqrt{2(1.6 \times 10^{-27} \; kg)(1.38 \times 10^{-23} \; J/K)(300 \;K)}} \tag{3}\]

And the wavelength is:

\[ \lambda = 0.18 nm \tag{4}\]

Q3

Calculate the wavelength of the highest energy Balmer emission line of hydrogen.

1. What is the angular momentum in the lowest energy state of the Bohr hydrogen atom?
2. What velocity of the electron in the lowest energy state

Solution

The Balmer series is:

\[ \bar{\nu}= \dfrac{1}{ \lambda} =R\left( \dfrac{1}{4} -\dfrac{1}{n_2^2}\right), n_2>2 \tag{1} \]

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

\[n_2 = \infty \tag{2}\]

Input the values:

\[ \dfrac{1}{ \lambda} =(1.09 \times 10^{7}\; 1/m)\left( \dfrac{1}{4} -\dfrac{1}{\infty}\right) \tag{3} \]

The wavelength is:

\[ \dfrac{1}{ \lambda} = 2725000\; 1/m \tag{4}\]

\[\lambda = 367 nm \tag{5}\]

In the Bohr atom, angular momentum is quantized

\[L = m_evr = n\hbar \tag{6}\]

And the lowest energy state is

\[n=1 \]

so:

\[L = \hbar \tag{\7}\]

which means that the velocity is:

\[v = \dfrac{\hbar}{m_er} \tag{8}\]

Input the values:

\[v = \dfrac{1.05 \times 10^{-34} \; J \cdot s}{(9.1 \times 10^{-31} \; kg)(5.29 \times 10^{-11} \; m)} \tag{9}\]

And the velocity is:

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

Q4

What is the minimum uncertainty in the speed of a 1 kg ball that is known to to be within \(1.0 \times 10^6\;m\) on a bat? What is the maximum uncertainly in the position of a 10 g carrot with a speed somewhere between 250.00001 m/s and 250.00000 m/s?

Solution

The Heisenberg Uncertainty Principle says:

\[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi} \tag{1}\]

If we insert the definition for momentum:

\[m\Delta{v}\Delta{x} \ge \dfrac{h}{4\pi} \tag{2}\]

\[\Delta{v} \ge \dfrac{h}{4m\pi\Delta{x}} \tag{3}\]

Now input the values for the baseball problem:

\[\Delta{v} \ge \dfrac{6.63 \times 10^{-34}\; J \cdot s}{4(1 \; kg)\pi(1.0 \times 10^6\;m)} \tag{4}\]

And

\[\Delta{v} \ge 5.3 \times 10^{-41} \; m/s \tag{5}\]

Note: I think the problem is meant to have \(\Delta{x} = 1 \times 10^{-6} \; m \) which would be more sensible for a baseball pitch. Then the answer is \(\Delta{v} \ge 5.3 \times 10^{-29} \; m/s \)

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.

\[\Delta{x} = \infty \tag{6}\]

Q5

The Heisenberg Uncertainty Principle says:

\[\Delta{p}\Delta{x} \ge \dfrac{h}{4\pi} \tag{1}\]

If we insert the definition for momentum:

\[m\Delta{v}\Delta{x} \ge \dfrac{h}{4\pi} \tag{2}\]

Rearrange for \(\Delta{x}\)

\[\Delta{x} \ge \dfrac{h}{4m\pi\Delta{v}} \tag{3}\]

Input the values in the correct units m/sec and kg for the really slow chicken.

\[\Delta{x} \ge \dfrac{h}{4 (0.1 \; kg) \pi\(1.0 \times 10^7\;m/s)} \tag{4}\]

\[\Delta{x} \ge 5.273 \times 10^{-27} \;m \tag{5}\]

Q6

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:

\[U(x,t) = e^{i(k x + ωt)} \tag{1}\]

And the classical wavefunction is:

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

\[k^2 e^{i(k x + ωt)} = -\dfrac{\omega^2}{v^2} e^{i(k x + ωt)} \tag{3}\]

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

(ii) The function is now:

\[U(x,t) = \cos(k\,x - \omega\, t) \tag{4}\]

\[\dfrac{\delta^{2}U}{\delta x^2} = \dfrac{1}{v^2} \dfrac{\delta^{2}U}{\delta t^2} \tag{5}\]

\[-k^2 \cos(k\,x - \omega\, t) = \dfrac{\omega^2}{v^2} \cos(k\,x - \omega\, t) \tag{6}\]

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

Q7

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

\[A \cos(\omega t) + B \sin(\omega t) = C \cos (\omega t + \Phi) = C \cos (\omega t) \cos (\Phi) - C \sin (\omega t) \sin (\Phi) \tag{1}\]

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

\[A\cos(\omega t) + B\sin(\omega t) = C\sin(\omega t + \Psi) = C\sin(\omega t)\cos(\Psi) – C\cos(\omega t)sin(\Psi) \tag{2}\]

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)

\[\dfrac{\delta U^2}{\delta t^2} + \omega^2 U = 0\tag{3}\]

\[U(t) = C\cos(\omega t + \Phi) \tag{4}\]

\[-\omega^2 C \cos(\omega t + \Phi) + \omega^2 C \cos (\omega t + \Phi) = 0 \tag{5} \]

(2.3)

\[\dfrac{\delta U^2}{\delta t^2} + \omega^2 U = 0\tag{6}\]

\[U(t) = C\sin(\omega t + \Psi) \tag{7}\]

\[-\omega^2 C \sin(\omega t + \Psi) + \omega^2 C \sin (\omega t + \Psi) = 0 \tag{8} \]

Q8

Wave equation between 0 and L

\[ \dfrac{\partial^2 u(x)}{\partial x^2} + \left( \dfrac{8\pi^2m E}{h^2} \right) u(x)= 0 \tag{1}\]

with these boundary conditions:

\[u(0)= u(L) = 0 \tag{2}\]

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

\[u(x) = A \sin kx + B \cos kx\tag{3}\]

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

\[k = \dfrac{nπ}{L}\tag{4}\]

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

\[ \dfrac{8π^2mE}{h^2} = \dfrac{n^2π^2}{L^2}\tag{5}\]

Then solving for \(E\) to get

\[E = \dfrac{n^2h^2}{8mL^2} \tag{6}\]

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