Exercises: Wave-particle duality: De Broglie and photoelectric effect

Light produces interference and diffraction patterns, like any wave, but when it hits a metal it knocks out electrons as if it were a stream of particles (photons).

1. a) Explain in your own words what the wave-particle duality of light consists of.
2. b) Name one phenomenon that can only be explained by wave behavior and another that can only be explained by particle behavior.
3. c) In $1924$, De Broglie extended the idea to matter: what hypothesis did he propose regarding electrons and other particles?

A dermatological laser emits light with frequency $f = 4.5\times10^{14}\ \text{Hz}$. The energy of a photon is $E_{fotón} = h\times f$, where $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$.

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: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$.

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

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 setup.

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.

Sodium vapor street lamps emit light at $\lambda_{Na} = 589\ \text{nm}$ and a blue LED emits at $\lambda_{az} = 470\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$. 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 photoelectric effect, each metal has a work function $W$ (also called the extraction work) and a threshold frequency $f_0$ satisfying $W = h\times f_0$.

1. a) Explain what the work function means: what energy does it represent?
2. b) What is the threshold frequency and what happens when the incident light has $f < f_0$?
3. c) If $f > f_0$, what happens to the remaining photon energy after the electron is emitted?

A fiber-optic laser for telecommunications emits at $\lambda = 1.55\ \mu\text{m}$ with a power of $10\ \text{mW}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$.

1. a) Calculate the energy of each photon.
2. b) Calculate how many photons the laser emits each second (remember that power is the energy emitted per unit time).

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: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. 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.

The light detector of a telescope uses a cesium cell, whose work function is $W = 2.1\ \text{eV}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

1. a) Calculate the threshold frequency of cesium, $f_0 = \frac{W}{h}$.
2. b) Calculate the threshold wavelength, $\lambda_0 = \frac{c}{f_0}$.
3. c) Will red light of $\lambda = 650\ \text{nm}$ emit electrons? Calculate this by comparing energies and explain your answer.

Light of $\lambda = 350\ \text{nm}$ is incident on a calcium plate with a work function of $W = 2.9\ \text{eV}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

1. a) Calculate the photon energy in joules and express it in electronvolts.
2. b) Determine whether photoelectron emission occurs.
3. c) If so, calculate the maximum kinetic energy of the emitted electrons.

The solar panel of a satellite receives $200\ \text{W}$ of light power from the Sun. As a simplification, assume that all the light has $\lambda = 550\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $c = 3\times10^{8}\ \text{m/s}$.

1. a) Calculate the energy of each photon.
2. b) Calculate how many photons reach the panel each second.
3. c) If with the same power the light were redder (larger $\lambda$), would more or fewer photons arrive per second? Justify your answer using the formula $E_{fotón} = \frac{h\times c}{\lambda}$.

A metal with work function $W = 2.25\ \text{eV}$ is illuminated with light of $\lambda = 300\ \text{nm}$. The photoelectrons are slowed down by a potential difference until even the fastest one does not reach the collector (stopping potential $V_f$, with $E_{c,máx} = e\times V_f$). Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$, $e = 1.6\times10^{-19}\ \text{C}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. The diagram shows the setup with the stopping potential.

1. a) Calculate the photon energy and the maximum kinetic energy of the photoelectrons, in electronvolts.
2. b) Calculate the stopping potential.
3. c) Repeat the calculation with light of $\lambda = 250\ \text{nm}$: how do the maximum kinetic energy and the stopping potential change?
4. d) Explain why doubling the light intensity, without changing its wavelength, does not change the stopping potential, even though it increases the number of photoelectrons.

A laboratory group measures the photoelectric effect on a metal and obtains two data points: with light of frequency $f_1 = 6.0\times10^{14}\ \text{Hz}$ there is no emission at all (it is the threshold frequency) and with $f_2 = 1.0\times10^{15}\ \text{Hz}$ photoelectrons leave with $E_{c,máx} = 2.65\times10^{-19}\ \text{J}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. The diagram shows the two data points from the experiment.

1. a) Calculate the work function of the metal from $W = h\times f_1$ and express it in electronvolts.
2. b) Mentally sketch $E_{c,máx}$ against $f$: using $E_{c,máx} = h\times f - W$, calculate the slope 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 $f = 1.2\times10^{15}\ \text{Hz}$.
4. d) Explain what quantity the y-intercept of the straight line represents and why it is negative.

A zinc plate, with a work function of $W = 4.3\ \text{eV}$, is illuminated alternately with red light of $650\ \text{nm}$ and ultraviolet light of $250\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

1. a) Calculate the energy of each type of photon and determine, with justification, which light causes emission.
2. b) For the light that produces emission, calculate the maximum kinetic energy of the photoelectrons and the corresponding stopping potential.
3. c) If we multiply the intensity of the red light by a thousand, will we succeed in knocking out electrons from the zinc? Explain your answer using the photon model: how do a thousand red photons differ from one ultraviolet photon?
4. d) If we increase the intensity of the ultraviolet light, what changes and what stays the same in the emission?

A photovoltaic cell receives $5\ \text{mW}$ of light of $\lambda = 600\ \text{nm}$. Given: $h = 6.63\times10^{-34}\ \text{J}\times\text{s}$, $c = 3\times10^{8}\ \text{m/s}$ and $e = 1.6\times10^{-19}\ \text{C}$.

1. a) Calculate how many photons reach the cell each second.
2. b) Assuming that each photon releases one electron that reaches the external circuit, calculate the maximum theoretical current (current is charge passing per second).
3. c) In practice the current is lower: give two reasons to explain this difference.
4. d) If the cell received the same power but from light of $\lambda = 400\ \text{nm}$, would the maximum theoretical current be greater, smaller, or equal? Justify your answer by counting photons.