\(ch = 1240 \text{eV nm}\)
For light of frequency \(f\) and wavelength \(\lambda\), the photon energy is:
\[ E = hf \]
Where:
de Broglie wavelength:
\[ \lambda = \frac{h}{p} \]
When light of high enough frequency illuminates a metal surface, the electrons of the metal surface can gain energy to escape the metal by absorbing the photons of the light. That’s called the photoelectric effect.
The conservation of energy of such an absorption and escape is written as:
\[ hf = K_{max} + \Phi \]
Where:
Momentum of a photon
\[ p = \frac{hf}{c} = \frac{h}{\lambda} \]
Wavelength increase due to Compton shift (where split x-rays lose energy to electrons of a target they hit)
\[ \Delta lambda = \frac{h}{mc} ( 1 - cos \phi ) \]
Where:
Spectral radiancy
\[ S( \lambda ) = \frac{\text{intensity}}{\text{(unit wavelength)}} \]
Planck radiation law:
\[ S( \lambda ) = \frac{2 \pi c^{2} h}{\lambda^{5}} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1} \]
Where:
Wien’s law. Relates the temperature \(T\) of a blackbody radiator and the wavelength \(\lambda_{max}\) at which the spectral radiancy is maximum.
\[ \lambda_{max} T = 2898 \mu m * K \]
A moving particle such as an electron can be described as a matter wave. The wavelength associated with the matter wave is the particle’s de Broglie wavelength:
\[ \lambda = \frac{h}{p} \]
Where:
\[ \frac{d^{2} \psi}{dx^{2}} + k^{2} \psi = 0 \]
Where:
\(k\) is related to \(\lambda\), which is the de Broglie wavelength, and the momentum \(p\), and the kinetic energy \(E - U\) by:
\[ k = \frac{2 \pi}{\lambda} = \frac{2 \pi p}{h} = \frac{2 \pi \sqrt{2m (E - U)}}{h} \]
A particle does not have a specific location until its location is actually measured
The probability of detecting a particle in a small volume cented on a given point is proportional to the probability density \(\lvert \psi \rvert^{2}\) of the matter wave at that point.
\[ \Delta x * \Delta p_{x} \geq \hbar \]
\[ \Delta y * \Delta p_{y} \geq \hbar \]
\[ \Delta z * \Delta p_{z} \geq \hbar \]
\(R\) is the average fraction of a beam with alot of particles that will undergo reflection
Transmission coefficient
\[ T = 1 - R \]
Transmission coefficient:
\[ T \approx e^{-2 b L} \]
Where \(b\) is:
\[ b = \sqrt{\frac{8 \pi^{2} m ( U_{b} - E )}{h^{2}}} \]
This question is regarding the photon.
At what rate does the Sun emit photons? For simplicity, assume that the Sun’s entire emission at the rate of \(3.9 \times 10^{26}\) W is at the single wavelength of 550 nm.
We know:
\[ P = R E \]
We have the power, and we need to find R (photons / s).
\[ R = \frac{P}{E} \]
\[ R = \frac{P}{\frac{hc}{\lambda}} \]
\[ R = \frac{P}{\frac{hc}{\lambda}} \]
I choose to stick to electron volts for this as I know \(hc = 1240\) eV nm and we are using nanometers.
\[ R = \frac{3.9 \times 10^{26}}{\frac{1240}{550}} * 6.24 \times 10^{18} \]
Our answer comes out to be \(1.0 \times 10^{45}\) photons/s, our answer.
This question is regarding the photon.
A light detector (your eye) has an area of \(2.00 \times 10^{-6}\) \(m^{2}\) and absorbs 80% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.00 m from the source. If the detector absorbs photons at the rate of exactly 4.000 \(s^{-1}\), at what power does the emitter emit light?
First, find the rate at which photons are being emitted from the light source:
\[ R_{detector} = (0.80) * \frac{A_{eye}}{4 \pi r^{2}} R_{source} \]
Find \(R_{source}\)
\[ R_{source} = \frac{4 \pi (3.00)^{2}}{(0.80) * 2 \times 10^{-6}} * 4 \]
\(R_{source}\) comes out to be \(2.83 \times 10^{8}\).
Next, find the energy of each photon. Once we do this, we know how much energy is contained per photon (energy/photon) and we have the rate at which photons are emitted (photons/s) so we can find the power (energy/s):
\[ E = \frac{hc}{\lambda} = \frac{1240}{540} \]
This comes out to be 2.48
Finally:
\[ P = RE = (2.83 \times 10^{8}) * (2.48) \]
This comes out to be \(1.1 \times 10^{-10}\) W, our answer.
This question is regarding the photoelectric effect.
You wish to pick an element for a photocell that will operate via the photoelectric effect with visible light. Which of the following are suitable (work functions are in parentheses): tantalum (4.2 eV), tungsten (4.5 eV), aluminum (4.2 eV), barium (2.5 eV), lithium (2.3 eV)?
First, find the maximum energy that visible light can provide us with:
\[ E = \frac{hc}{\lambda} \]
\[ E = \frac{6.626 \times 10^{-34} * (3 \times 10^{8})}{400} \]
We choose 400 here (lower bound of the range for visible light) to maximize the energy, so we have a better opportunity to satisfy the work function. We get \(3.1\) eV as the energy of the absorbed photon.
Thus, we would choose barium and lithium as our materials, since the maximum energy exceeds these work functions.
What percentage increase in wavelength leads to a 75% loss of photon energy in a photon-free electron collision?
We know:
\[ E - E' = E - 0.25 E \]
\[ E - 0.25 E = \frac{hc}{\lambda} - \frac{hc}{\lambda^{'}} \]
\[ 0.75 E = \frac{hc}{\lambda} - \frac{hc}{\lambda^{'}} \]
\[ \frac{0.75 E}{h} = \frac{c}{\lambda} - \frac{c}{\lambda^{'}} \]
We know \(E = \frac{hc}{\lambda}\)
\[ \frac{0.75c}{\lambda} = \frac{c}{\lambda} - \frac{c}{\lambda^{'}} \]
\[ 0.75 = 1 - \frac{\lambda}{\lambda^{'}} \]
\[ 0.25 = \frac{\lambda}{\lambda^{'}} \]
\[ \lambda^{'} = 4 \lambda \]
So:
\[ \lambda^{'} - \lambda = 4 \lambda - \lambda = 3 \lambda \]
Thus, a 300% increase leads to a 75% loss of photon energy in a photon-free electron collision.
Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{7}\) K.
- Find the wavelength at which the thermal radiation is maximum
- Identify the type of electromagnetic wave corresponding to that wavelength. (See Fig. 33-1.)
- This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{5}\) K. Find the wavelength at which the thermal radiation is maximum.
- Identify the type of electromagnetic wave corresponding to that wavelength.
For this question, we use Wien’s Law:
\[ \lambda_{max} T = 2898 \mu m \]
Solve for \(\lambda_{max}\) for part a:
\[ \lambda_{max} = \frac{2898}{1 \times 10^{7}} \]
This comes out to be \(2.9 \times 10^{-10}\) m, our answer.
From the electromagnetic wave spectrum we see that x-rays correspond to this wavelength.
For part c and d, we repeat the procedure using a different value for the temperature:
\[ \lambda_{max} = \frac{2898}{1 \times 10^{5}} \]
This comes out to be \(2.9 \times 10^{-8}\) m, our answer.
From the electromagnetic wave spectrum we see that ultraviolet waves correspond to this wavelength.
This problem is regarding quantum physics.
The wavelength of the yellow spectral emission line of sodium is 590 nm. At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?
\[ E = \frac{1}{2} m v^{2} \]
\[ E = \frac{1}{2} \frac{p^{2}}{m} = \frac{p^{2}}{2m} \]
Solve for momentum:
\[ p = \sqrt{2mE} \]
de Broglie wavelength:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \]
Now, solve for \(E\):
\[ \lambda^{2} = \frac{h^{2}}{2mE} \]
\[ E = \frac{h^{2}}{2m\lambda^{2}} \]
\[ E = \frac{(6.626 \times 10^{-34})^{2}}{2 * 9.11 \times 10^{-31} * (590 \times 10^{-9})^{2}} \]
\[ E = 6.92 \times 10^{-25} J \]
Convert \(J\) to eV via conversion factor:
\[ \frac{6.92 \times 10^{-25}}{1.602 \times 10^{-19}} \]
Thus, we get our answer of \(4.3 \mu\) eV.
The function \(\psi (x)\) displayed in Eq. 38-27 can describe a free particle, for which the potential energy is U(x) = 0 in Schrodinger’s equation (Eq. 38-19). Assume now that \(U(x) = U_{0}\) = a constant in that equation. Show that Eq. 38-27 is a solution of Schrodinger’s equation, with:
\[ k = \frac{2 \pi}{h} \sqrt{2m ( E - U_{0} )} \]
giving the angular wave k of the particle.
\(E - U_{0}\) represents the kinetic energy. So, we know the kinetic energy can be expressed as:
\[ K = \frac{p^{2}}{2m} \]
Momentum in terms of the kinetic energy is then:
\[ p = \sqrt{2m K} \]
Momentum in terms of the de Broglie wavelength:
\[ p = \frac{h}{\lambda} \]
Now we can represent \(k\) in terms of the momentum:
\[ k = \frac{2 \pi}{\lambda} = \frac{p 2 \pi}{h} \]
Substitute \(p\) with its representation in terms of the kinetic energy now:
\[ k = \frac{2 \pi \sqrt{2mK}}{h} \]
Thus, Eq. 38-27 is a solution to Schrodinger’s Equation.
This question is regarding Heisenburg’s uncertainty principle.
The uncertainty in the position of an electron along an x axis is given as 50 pm, which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of the momentum component \(p_{x}\) of this electron?
We use Heisenburg’s uncertainty principle:
\[ \Delta p_{x} = \frac{\hbar}{\Delta x} \]
\[ \Delta p_{x} = \frac{\frac{6.626 \times 10^{-34}}{2 \pi}}{50 \times 10^{-12}} \]
\[ \Delta p_{x} = 2.1 \times 10^{-24} \]
Thus, \(\Delta p_{x} = 2.1 \times 10^{-24}\) kg m/s is our answer.
This question is regarding reflection through a potential step.
For the arrangement of Figs. 38-14 and 38-15, electrons in the incident beam in region 1 have energy E = 800 eV and the potential step has a height of \(U_{1}\) = 600 eV. What is the angular wave number in
- region 1
- region 2
Also,
- What is the reflection coefficient?
- If the incident beam sends \(5.00 \times 10^{5}\) electrons against the potential step, approximately how many will be reflected?
Region 1 (use 800 eV since U = 0):
\[ k = \frac{2 \pi \sqrt{2 m E}}{h} \]
\[ k = \frac{2 \pi \sqrt{2 (9.11 \times 10^{-31}) (800 * (1.602 \times 10^{-19}))}}{6.626 \times 10^{-34}} \]
Thus, we get \(1.45 \times 10^{11} m^{-1}\) as our answer.
Now, for Region 2, there is U = 600, so use 200eV:
\[ k = \frac{2 \pi \sqrt{2 m (E - U_{1})}}{h} \]
\[ k = \frac{2 \pi \sqrt{2 (9.11 \times 10^{-31}) ((800 - 600) * (1.602 \times 10^{-19}))}}{6.626 \times 10^{-34}} \]
Thus, we get \(7.25 \times 10^{10} m^{-1}\) as our answer.
We know the reflection coefficient can be expressed as:
\[ R = \frac{\lvert B \rvert^{2}}{\lvert A \rvert^{2}} \]
We know from Equations 38-34 and 38-35:
\[ A + B = C \]
\[ A * k - B * k = C * k_{b} \]
Considering we know that \(k_{b} = \frac{k}{2}\) (this is because 800 was bigger by a factor of 4 to 200, and it was in the square root), let us perform a substitution into the latter equation.
\[ A * k - B * k = C * \frac{k}{2} \]
\[ A - B = \frac{C}{2} \]
\[ A - B = \frac{C}{2} \]
Let us add \(2B\) to each side:
\[ A + B = \frac{C}{2} + 2B \]
Now we can equate the two equations via transitivity:
\[ \frac{C}{2} + 2B = C \]
\[ \frac{C}{2} + 2B = C \]
\[ 2B - \frac{C}{2} = 0 \]
\[ 2B = \frac{C}{2} \]
\[ 4 = \frac{C}{B} \]
Now let us represent \(B\) in terms of \(A\):
\[ 4 = \frac{C}{C - A} \]
\[ 4C - 4A = C \]
\[ 3C - 4A = 0 \]
\[ 3C = 4A \]
\[ \frac{3}{4} = \frac{A}{C} \]
Considering we have found \(\frac{C}{B}\) and \(\frac{A}{C}\), we multiply to find \(\frac{A}{B}\):
\[ 4 * \frac{3}{4} = 3 \]
Inverse this to find \(\frac{B}{A}\):
\[ 3^{-1} = \frac{1}{3} \]
Square it to find \(\frac{\lvert B \rvert^{2}}{\lvert A \rvert^{2}}\), which is \(R\):
\[ ( \frac{1}{3} )^{2} = \frac{1}{9} \]
Thus, \(\frac{1}{9}\) is our answer.
Now, to find how many electrons will be reflected from an incident beam that sends \(5 \times 10^{5}\) electrons, we multiply by the reflection coefficient for our answer:
\[ 5 \times 10^{5} * \frac{1}{9} \]
Thus, our answer is \(5.56 \times 10^{4}\) electrons.
This question is regarding tunneling through a potential barrier.
Consider a potential energy barrier like that of Fig. 38‑17 but whose height \(U_{b}\) is 6.0 eV and whose thickness L is 0.70 nm. What is the energy of an incident electron whose transmission coefficient is 0.0010?
We know:
\[ T \approx e^{-2 b L} \]
and \(b\) is:
\[ b = \sqrt{\frac{8 \pi^{2} m ( U_{b} - E )}{h^{2}}} \]
So, solve for \(E\):
\[ ln T = -2 b L \]
\[ \frac{(ln T)}{-2L} = b \]
\[ \frac{(ln T)}{-2L} = \sqrt{\frac{8 \pi^{2} m ( U_{b} - E )}{h^{2}}} \]
\[ (\frac{(ln T)}{-2L})^{2} = \frac{8 \pi^{2} m ( U_{b} - E )}{h^{2}} \]
\[ (\frac{(h ln T)}{\pi -2 L})^{2} = 8 m ( U_{b} - E ) \]
\[ E = U - \frac{((\frac{(h ln T)}{\pi -2 L})^{2})}{8m} \]
\[ E = (6.0 \times 1.602 \times 10^{-19}) - \frac{((\frac{((6.626 \times 10^{-34}) * ln 0.0010)}{\pi -2 * (0.70 \times 10^{-9})})^{2})}{8 * (9.11 \times 10^{-31})} \]
Substituting values, we get \(E = 5.1\) eV, our answer.
This question is regarding tunneling through a potential barrier.
An electron with total energy E = 5.1 eV approaches a barrier of height U b = 6.8 eV and thickness L = 750 pm. What percentage change in the transmission coefficient T occurs for a 1.0% change in
- the barrier height
- the barrier thickness
- the kinetic energy of the incident electron?
-17%
First let us calculate \(T\) without any variations:
\[ e^{-2 * (750 \times 10^{-12}) * \sqrt{\frac{8 * \pi^{2} * (9.11 \times 10^{-31}) * ((6.8 - 5.1) * 1.602 \times 10^{-19})}{(6.626 \times 10^{-34})^{2}}}} \]
We can then use this as a reference value (i.e. I stored this value into my calculator’s memory var \(A\)) to solve for the percent change when parameters are varied by 1.0%.
When the barrier height is changed by 1%, new value of transmission coefficient (say it is \(B\)) is:
\[ e^{-2 * (750 \times 10^{-12}) * \sqrt{\frac{8 * \pi^{2} * (9.11 \times 10^{-31}) * (((1.01 * 6.8) - 5.1) * 1.602 \times 10^{-19})}{(6.626 \times 10^{-34})^{2}}}} \]
Calculate \(\frac{100 * (B - A)}{A}\) for percent change:
Thus our answer is -18%.
When the barrier thickness is changed by 1%, new value of transmission coefficient (say it is \(B\)) is:
\[ e^{-2 * (1.01 * 750 \times 10^{-12}) * \sqrt{\frac{8 * \pi^{2} * (9.11 \times 10^{-31}) * ((6.8 - 5.1) * 1.602 \times 10^{-19})}{(6.626 \times 10^{-34})^{2}}}} \]
Calculate \(\frac{100 * (B - A)}{A}\) for percent change:
Thus our answer is -9.5%.
When the kinetic energy of the incident electron is changed by 1%, new value of transmission coefficient (say it is \(B\)) is:
\[ e^{-2 * (750 \times 10^{-12}) * \sqrt{\frac{8 * \pi^{2} * (9.11 \times 10^{-31}) * ((6.8 - (1.01 * 5.1)) * 1.602 \times 10^{-19})}{(6.626 \times 10^{-34})^{2}}}} \]
Calculate \(\frac{100 * (B - A)}{A}\) for percent change:
Thus our answer is 16%.