Wave-particle duality: De Broglie and the photoelectric effect

A dermatological laser emits light with frequency $f = 4,5\cdot10^{14}\ \text{Hz}$. The energy of a photon is $E_{fotón} = h\cdot f$, where $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$.

1. a) Calculate the energy of each photon.
2. b) If we increase the intensity of the laser without changing its frequency, does the energy of each photon increase? What actually increases?

A laser pointer emits green light with wavelength $\lambda = 550\ \text{nm}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $c = 3\cdot10^{8}\ \text{m/s}$.

1. a) Calculate its frequency, $f = \frac{c}{\lambda}$.
2. b) Calculate the energy of each photon, $E_{fotón} = h\cdot f$.

Sodium vapor street lamps emit light at $\lambda_{Na} = 589\ \text{nm}$ and a blue LED emits at $\lambda_{az} = 470\ \text{nm}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $c = 3\cdot10^{8}\ \text{m/s}$. The diagram compares the two waves.

1. a) Calculate the energy of a photon of each type of light.
2. b) Which of the two photons is more energetic? How many times more?

In the laboratory, a negatively charged zinc plate is connected to an electroscope. When illuminated with ultraviolet light, the electroscope discharges; with very intense red light, it does not. The diagram shows the experimental set-up.

1. a) What is the photoelectric effect?
2. b) What are photoelectrons?
3. c) Explain, using the concept that light arrives in photons, why red light does not discharge the plate no matter how intense it is.

A potassium photocell has a work function of $W = 2,3\ \text{eV}$ and is illuminated with ultraviolet light of $\lambda = 250\ \text{nm}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$, $c = 3\cdot10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$. The diagram shows the schematic of the process.

1. a) Express the work function in joules.
2. b) Calculate the energy of each photon of ultraviolet light.
3. c) Apply $E_{fotón} = W + E_{c,máx}$ and calculate the maximum kinetic energy of the photoelectrons, in joules and in electronvolts.

A laboratory group measures the photoelectric effect on a metal and obtains two data points: with light of frequency $f_1 = 6,0\cdot10^{14}\ \text{Hz}$ there is no emission at all (it is the threshold frequency) and with $f_2 = 1,0\cdot10^{15}\ \text{Hz}$ photoelectrons leave with $E_{c,máx} = 2,65\cdot10^{-19}\ \text{J}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$. The diagram shows the two data points from the experiment.

1. a) Calculate the work function of the metal from $W = h\cdot f_1$ and express it in electronvolts.
2. b) Mentally sketch $E_{c,máx}$ against $f$: using $E_{c,máx} = h\cdot f - W$, calculate the gradient of the straight line through the two points and check that it matches the value of $h$.
3. c) Predict, without measuring again, the maximum kinetic energy for light of $f = 1,2\cdot10^{15}\ \text{Hz}$.
4. d) Explain what quantity the y-intercept of the straight line represents and why it is negative.

Every particle has an associated De Broglie wavelength, $\lambda = \frac{h}{m\cdot v}$, where $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$. A tennis ball of mass $58\ \text{g}$ leaves a serve at $40\ \text{m/s}$.

1. a) Calculate its De Broglie wavelength.
2. b) Compare the result with the size of an atom ($10^{-10}\ \text{m}$) and explain why this wave has no observable consequences.

In a cathode-ray tube, electrons move at $v = 10^{6}\ \text{m/s}$. Given: $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and mass of an electron $m_e = 9,11\cdot10^{-31}\ \text{kg}$.

1. a) Calculate the De Broglie wavelength associated with each electron, $\lambda = \frac{h}{m_e\cdot v}$.
2. b) Compare the result with the typical size of an atom ($10^{-10}\ \text{m}$) and comment on why the wave associated with the electron is relevant in the atomic world.

In an accelerator, an electron and a proton move at the same speed, $v = 10^{5}\ \text{m/s}$. Given: $m_e = 9,11\cdot10^{-31}\ \text{kg}$, $m_p = 1,67\cdot10^{-27}\ \text{kg}$ and $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$.

1. a) Calculate the De Broglie wavelength of the electron.
2. b) Calculate that of the proton.
3. c) Calculate the ratio $\frac{\lambda_e}{\lambda_p}$ and explain, using De Broglie's formula, why at the same speed the more massive particle has a shorter associated wavelength.

In a transmission electron microscope, electrons are accelerated through a potential difference $V = 10\ \text{kV}$, gaining a kinetic energy $E_c = e\cdot V$ (ignore relativistic corrections). Given: $e = 1,6\cdot10^{-19}\ \text{C}$, $m_e = 9,11\cdot10^{-31}\ \text{kg}$ and $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$.

1. a) Calculate the kinetic energy acquired by each electron, in joules.
2. b) Calculate its speed.
3. c) Calculate its De Broglie wavelength.
4. d) The resolving power of a microscope improves as the wavelength used decreases: compare the wavelength obtained with that of visible light ($\lambda \approx 500\ \text{nm}$) and explain why an electron microscope can resolve viruses while an optical microscope cannot.

For a certain diffraction experiment, the electrons must have a De Broglie wavelength of $1\ \text{nm}$. Given: $m_e = 9,11\cdot10^{-31}\ \text{kg}$, $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$ and $1\ \text{eV} = 1,6\cdot10^{-19}\ \text{J}$.

1. a) Rearrange $\lambda = \frac{h}{m_e\cdot v}$ for speed and calculate the speed at which they must move.
2. b) Calculate their kinetic energy, in joules and in electronvolts.

In a double-slit experiment with electrons, the electrons move at $v = 2\cdot10^{6}\ \text{m/s}$, the slits are separated by $d = 0,5\ \mu\text{m}$, and the screen is at $L = 1\ \text{m}$. The separation between consecutive maxima in the pattern is $\Delta y = \frac{\lambda\cdot L}{d}$. Given: $m_e = 9,11\cdot10^{-31}\ \text{kg}$ and $h = 6,63\cdot10^{-34}\ \text{J}\cdot\text{s}$. The diagram shows the double-slit set-up.

1. a) Calculate the De Broglie wavelength of the electrons.
2. b) Calculate the separation between maxima on the screen and express it in millimetres.
3. c) If electrons are sent one by one, the interference pattern still forms over time: explain what this result means regarding wave-particle duality and the probabilistic nature of quantum physics.
4. d) Predict what happens to the separation between maxima if the electrons are accelerated to twice the speed.