Photons

(b) $E^2 = P^2C^2$

A photon has zero rest mass, so the relativistic relation $E^2 = p^2c^2 + m_0^2c^4$ reduces to $E = pc$. Squaring gives $E^2 = p^2c^2$.

(d) Graph 4: a straight line that cuts the positive $f$-axis, with a dashed extension to a negative intercept on the $V_0$-axis

From Einstein's equation, $eV_0 = hf - \phi$, so $V_0 = \dfrac{h}{e}f - \dfrac{\phi}{e}$. This is a straight line with slope $h/e$. It is zero at the threshold frequency $f_0 = \phi/h > 0$, and its extension meets the $V_0$-axis at $-\phi/e$.

(a) Work function

The work function $\phi$ of a metal is the minimum energy needed to remove an electron from the surface of the metal. Electrons inside the metal are held by the attraction of the positive ions, and at least this much energy must be supplied to free one. It depends on the nature of the metal. If light of frequency $f$ falls on it, electrons are emitted only if $hf \ge \phi$. The threshold frequency is $f_0 = \phi/h$.

(b) Will photoelectric emission occur?

Given: $\phi = 6.4\times10^{-19}$ J, $f = 1.2\times10^{15}$ Hz, $h = 6.62\times10^{-34}$ J s

Since $E = 7.94\times10^{-19}$ J $> \phi = 6.4\times10^{-19}$ J, yes, the photoelectric effect will take place.

(Equivalently, the threshold frequency is $f_0 = \phi/h = 9.67\times10^{14}$ Hz, which is less than $1.2\times10^{15}$ Hz.)

(c) (i) Einstein's photoelectric equation

(c) (ii) Extra energy available

Given: $f = 1.5\times10^{15}$ Hz

This extra energy becomes the maximum kinetic energy of the emitted electron.

(c) (iii) Highest-energy photons

The blue lamp. Photon energy is $E = hf = hc/\lambda$. Blue light has the highest frequency (shortest wavelength) of the three colours, so its photons have the most energy. The order is red < green < blue.

   V0
    ^                 /
    |               /
    |             /   slope = h/e
    |           /
  O +---------*-----------> f
    |       / f0
    |     /  (dashed)
    |   /
    | /
    * -phi/e

i) Einstein's equation in terms of stopping potential

ii) Finding Planck's constant from the graph

The equation $V_0 = \left(\dfrac{h}{e}\right)f - \dfrac{\phi}{e}$ has the form $y = mx + c$, so the graph is a straight line with slope $= h/e$.

1. Take two points on the line, $(f_1, V_1)$ and $(f_2, V_2)$, and find the slope $= \dfrac{V_2 - V_1}{f_2 - f_1} = \dfrac{\Delta V_0}{\Delta f}$.
2. Multiply the slope by the electronic charge:

(The intercept on the $f$-axis gives the threshold frequency $f_0$. The intercept on the $V_0$-axis gives $-\phi/e$, from which $\phi$ can be found.)

iii) Effect of gradually increasing the frequency

The stopping potential increases linearly with frequency, since $V_0 = \dfrac{h}{e}f - \dfrac{\phi}{e}$, with the work function $\phi$ fixed for a given metal. A higher frequency means each photon has more energy $hf$. After the fixed work function is paid, the photoelectrons have more maximum kinetic energy, $\tfrac12mv_{max}^2 = hf - \phi$, so a larger retarding potential is needed to stop them. (Below $f_0$ there is no emission, so $V_0 = 0$ there.)

Photoelectric effect: When light of a suitably high frequency (above the threshold frequency $f_0$) falls on a metal surface, electrons are emitted from the surface. This is called the photoelectric effect, and the emitted electrons are called photoelectrons. The emission is instantaneous. The maximum kinetic energy of the photoelectrons depends on the frequency of the light, not on its intensity.

Einstein's photoelectric equation: Einstein treated light as a stream of photons, each of energy $hf$. A photon gives all its energy to one electron. Part of this energy, equal to the work function $\phi$, is used to free the electron, and the rest becomes its kinetic energy:

The threshold wavelength $\lambda_0$ of a metal is the maximum wavelength of incident light that can just eject electrons from its surface. For light of wavelength longer than $\lambda_0$, no photoelectric emission occurs, however intense the light. It is related to the work function by $\phi = \dfrac{hc}{\lambda_0}$, that is, $\lambda_0 = \dfrac{hc}{\phi}$.

Given: $\phi = 2$ eV $= 2\times1.6\times10^{-19} = 3.2\times10^{-19}$ J, $\lambda = 150$ nm $= 1.5\times10^{-7}$ m, $h = 6.62\times10^{-34}$ J s, $m = 9.1\times10^{-31}$ kg, $c = 3\times10^{8}$ m s$^{-1}$

1. Energy of an incident photon

2. Maximum kinetic energy (Einstein's equation)

3. Maximum speed

4. Least (threshold) frequency

The photoelectric effect is the emission of electrons from a metal surface when light (electromagnetic radiation) of frequency greater than a certain threshold frequency falls on it. The emitted electrons are called photoelectrons.

Einstein explained the photoelectric effect using Planck's quantum idea. Light consists of packets of energy called photons, each of energy $E = hf$.

• When a photon falls on a metal, it gives all its energy to one electron, in a single interaction.
• Part of this energy is used to free the electron from the surface. This minimum energy is the work function $\phi = hf_0$.
• The remaining energy appears as the maximum kinetic energy of the emitted electron.

Conclusions from the equation:

1. If $f < f_0$, the kinetic energy would be negative, so no electrons are emitted. This explains the threshold frequency.
2. $K_{max}$ increases linearly with $f$ and does not depend on intensity.
3. A higher intensity means more photons per second, so more electrons are emitted (a larger photocurrent), but $K_{max}$ is the same.
4. A single photon is absorbed in a single event, so the emission is instantaneous.

From Einstein's equation, $K_{max} = hf - \phi$. So the graph of $K_{max}$ against $f$ is a straight line whose slope is $h$ and whose intercept on the $K_{max}$-axis is $-\phi$.

Readings from the graph:

• Point A: $f_0 = 10\times10^{14}$ Hz $= 1.0\times10^{15}$ Hz, $K_{max} = 0$
• Upper end: $f = 30\times10^{14}$ Hz $= 3.0\times10^{15}$ Hz, $K_{max} = 8$ eV
• Point C (intercept on the $K_{max}$-axis): $-4$ eV

Planck's constant (slope):

Work function:

The intercept OC $= -\phi = -4$ eV, so

Check: $\phi = hf_0 = 4\times10^{-15}\ \text{eV s}\times1.0\times10^{15}\ \text{Hz} = 4$ eV. This agrees.

From Einstein's equation, $E_k = hf - \phi$, so the graph of $E_k$ against $f$ is a straight line with slope $h$, intercept on the $f$ axis equal to $f_0$ and intercept on the $E_k$ axis equal to $-\phi$.

(i) Threshold frequency

The line cuts the $f$ axis ($E_k = 0$) at $f_0 = 3\times10^{14}$ Hz.

(ii) Planck's constant

The intercept on the $E_k$ axis is $-2\times10^{-19}$ J, so $\phi = 2\times10^{-19}$ J. The slope is

(iii) Work function

(a) The photoelectric effect is the emission of electrons from a metal surface when light of suitable frequency (above a threshold frequency) falls on it.

(b) No. A photon can eject an electron only if its energy $hf$ is at least the work function $\phi$ of the material, i.e. $f\ge f_0$. Photons of lower frequency do not have enough energy, and no electron is emitted however intense the light is, because one photon gives its energy to one electron.

(c) To eject electrons using any visible light, even the weakest photons (red, 700 nm) must be able to do so, so the work function must not exceed the energy of a 700 nm photon:

$\phi \le \frac{hc}{\lambda} = \frac{1240\ \text{eV nm}}{700\ \text{nm}} \approx 1.77\ \text{eV}$

(The 400 nm violet photon has $1240/400 = 3.1$ eV.) So the work function must be about 1.77 eV or less for the whole visible range to eject photoelectrons.

From Einstein's equation, $E_{k,max} = hf - \phi$. So the graph is a straight line of slope $h$ that cuts the $f$-axis at the threshold frequency $f_0$.

(a) Threshold frequency

The line meets the frequency axis ($E = 0$) at about

This is the minimum frequency that can cause photoelectric emission from sodium. Any frequency above it also causes emission, so there is no "maximum" frequency; the question means the threshold value.

(b) Work function of sodium

Slope of the line, using the points $(5.6\times10^{14}\ \text{Hz}, 0)$ and $(8.0\times10^{14}\ \text{Hz}, 1.0\ \text{eV})$:

Extend the line back to $f = 0$. Its intercept on the $E$-axis is $-\phi$:

(c) Planck's constant

The slope of the graph equals $h$:

This is close to the accepted value, $6.63\times10^{-34}$ J s. Small differences come from reading the graph.

(i) Maximum speed of the photoelectrons

Given: $f = 1.5\times10^{15}$ Hz, $\phi = 6.6\times10^{-19}$ J, $h = 6.63\times10^{-34}$ J s, $m = 9.11\times10^{-31}$ kg

1. Photon energy:

2. Maximum kinetic energy:

3. Maximum speed:

This is the value we had to show.

(ii) Effect of increasing the intensity

There is no change in the maximum speed.

• Increasing the intensity (at the same frequency) increases the number of photons striking the surface per second. The energy of each photon, $hf$, stays the same.
• Each electron absorbs only one photon, so its maximum kinetic energy, $hf - \phi$, and therefore $v_{max}$, is unchanged.
• What does change is the number of electrons emitted per second: the photoelectric current increases in proportion to the intensity.