#2 Solutions
- Page ID
- 92308
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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.
- What is the angular momentum in the lowest energy state of the Bohr hydrogen atom?
- 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, …