Dual Nature Of Radiation And MatterClass 12 Physics Notes

Introduction

At the end of the 19th century, the scientific community was confident that light behaved as a wave, thanks to Maxwell's equations and Hertz's experiments with electromagnetic waves. At the same time, experiments with electricity passing through gases at low pressures led to groundbreaking discoveries like X-rays and, most importantly, the electron by J. J. Thomson in 1897.

Thomson studied cathode rays, which were streams of particles moving from the negative electrode (cathode) in a discharge tube. He found that these particles were negatively charged and determined their charge-to-mass ratio ($e/m$). A key finding was that this ratio was the same regardless of the gas in the tube or the metal used for the cathode. This suggested that these particles, which he named electrons, were a fundamental and universal part of all matter.

Around the same time, scientists noticed that certain metals emitted these same negatively charged particles when illuminated with ultraviolet light or when heated to high temperatures. This confirmed that electrons are universal constituents of matter.

Electron Emission

Metals contain free electrons that can move around inside the material, which is why metals are good conductors of electricity. However, these electrons don't normally escape the metal surface. If an electron tries to leave, the surface becomes positively charged and pulls it back.

To escape, an electron needs to be given enough energy to overcome this attractive force. The minimum amount of energy required for an electron to escape from a metal surface is called the work function of the metal, denoted by $\phi_0$. The work function is a property of the specific metal and its surface condition. It's often measured in electron volts (eV).

There are three main ways to supply this energy to electrons for emission:

• Thermionic emission: Heating the metal gives the free electrons enough thermal energy to escape.
• Field emission: Applying a very strong electric field (around $10^8 \text{ V m}^{-1}$) to the metal can pull the electrons out.
• Photoelectric emission: When light of a suitable frequency shines on a metal surface, electrons are emitted. These electrons are called photoelectrons.

Photoelectric Effect

The photoelectric effect is the phenomenon of electrons being emitted from a metal surface when light shines on it. It was first observed by Heinrich Hertz in 1887.

Hertz's Observations

Hertz noticed that sparks jumped more easily across a gap in his detector when the emitter plate was illuminated by ultraviolet (UV) light. This suggested that light was helping charged particles escape the metal surface.

Hallwachs' and Lenard's Observations

Wilhelm Hallwachs and Philipp Lenard studied this effect in more detail.

• Lenard found that when UV light hit an emitter plate in an evacuated glass tube, a current flowed. The current stopped as soon as the light was turned off. This showed that light was causing electrons to be ejected from the emitter.
• Hallwachs observed that a negatively charged zinc plate lost its charge when exposed to UV light. An uncharged zinc plate became positively charged. This confirmed that negatively charged particles (electrons) were being emitted.

A crucial observation they made was that for each metal, there is a certain minimum frequency of light, called the threshold frequency ($v_0$), below which no electrons are emitted, no matter how bright the light is.

• Metals like zinc and cadmium only respond to ultraviolet light.
• Alkali metals like sodium and potassium can emit electrons even with visible light.

Experimental Study of Photoelectric Effect

The setup to study the photoelectric effect (see Figure 11.1 in your textbook) consists of an evacuated tube with two metal plates: an emitter (C) and a collector (A). Light of a specific frequency and intensity shines on the emitter plate C, causing it to emit photoelectrons. A battery creates a potential difference between the plates, which can be varied or even reversed. The current produced by the flow of photoelectrons is measured with a microammeter.

Effect of intensity of light on photocurrent

With the collector plate A held at a positive potential and the light frequency fixed, we can vary the intensity (brightness) of the light.

• Observation: The photoelectric current is directly proportional to the intensity of the incident light.
• Conclusion: This means that a brighter light causes more photoelectrons to be emitted per second.

Effect of potential on photoelectric current

With the light frequency and intensity fixed, we can vary the potential of the collector plate A.

• Accelerating Potential: When A is positive relative to C, it attracts the emitted electrons. As the positive potential increases, the photocurrent increases until it reaches a maximum value called the saturation current. At this point, all emitted electrons are being collected by plate A.
• Retarding Potential: When A is made negative relative to C, it repels the electrons. Only the most energetic electrons can reach the collector. As the negative potential increases, the current decreases, eventually dropping to zero.

The minimum negative potential ($V_0$) at which the photocurrent becomes zero is called the stopping potential or cut-off potential. This potential is just enough to stop even the most energetic photoelectrons. The maximum kinetic energy ($K_{\max}$) of a photoelectron is therefore related to the stopping potential by: $K_{\max} = eV_0$

If we repeat the experiment with higher intensity light (but the same frequency), the saturation current increases (more electrons are emitted), but the stopping potential ($V_0$) remains the same.

• Conclusion: The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.

Effect of frequency of incident radiation on stopping potential

Now, we keep the intensity constant but vary the frequency of the incident light.

• Observation: The stopping potential ($V_0$) becomes more negative as the frequency of the light increases. The saturation current, however, remains the same.
• Conclusion: This implies that the maximum kinetic energy of the photoelectrons increases with the frequency of the incident light.

If we plot a graph of stopping potential ($V_0$) versus frequency ($\nu$), we get a straight line. This graph shows two important things:

1. The stopping potential $V_0$ (and thus $K_{\max}$) increases linearly with the frequency $\nu$.
2. There is a minimum cut-off frequency, the threshold frequency ($\nu_0$), below which the stopping potential is zero and no photoelectric emission occurs, regardless of the light's intensity.

Summary of Experimental Observations

1. Intensity: For a given frequency, the photocurrent is directly proportional to the intensity of light.
2. Stopping Potential: For a given frequency, the stopping potential is independent of intensity.
3. Frequency: There is a threshold frequency ($\nu_0$) below which no photoelectrons are emitted. Above this frequency, the stopping potential (and $K_{\max}$) increases linearly with the frequency of the incident light.
4. Instantaneous Process: Photoelectric emission is nearly instantaneous (occurs in less than $10^{-9}$ s), even for very dim light.

Photoelectric Effect and Wave Theory of Light

The classical wave theory of light, which was well-established, could not explain the observations of the photoelectric effect. Here’s why:

• Intensity Problem: According to wave theory, a more intense (brighter) light wave has more energy. Therefore, it should give electrons more kinetic energy. However, experiments show that the maximum kinetic energy depends only on the light's frequency, not its intensity.
• Frequency Problem: Wave theory predicts that any frequency of light, if intense enough, should eventually provide an electron with enough energy to escape the metal. This means there should be no threshold frequency. But experiments clearly show a threshold frequency exists.
• Time Lag Problem: In the wave picture, energy is spread out over the wavefront. An electron would need time (from minutes to hours for dim light) to absorb enough energy to be ejected. Experiments show that emission is instantaneous.

Because of these direct contradictions, it was clear that the wave theory of light was incomplete and could not explain the photoelectric effect.

Einstein's Photoelectric Equation: Energy Quantum of Radiation

In 1905, Albert Einstein proposed a revolutionary idea to explain the photoelectric effect. He suggested that light is not a continuous wave but is made up of discrete packets of energy called quanta (later named photons).

The energy of a single photon is directly proportional to the frequency of the light: $E = h\nu$ where $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J s}$).

In Einstein's picture, the photoelectric effect happens when one electron absorbs one photon.

1. An electron absorbs the entire energy ($h\nu$) from a single photon.
2. A part of this energy is used to overcome the attractive forces of the metal. This is the work function, $\phi_0$.
3. The rest of the energy becomes the kinetic energy of the electron.

The electron with the maximum kinetic energy ($K_{\max}$) is the one that uses the minimum energy to escape (the work function). This leads to Einstein's photoelectric equation: $K_{\max} = h\nu - \phi_0$

This simple equation elegantly explains all the experimental observations:

• Dependence on Frequency: The equation shows that $K_{\max}$ increases linearly with frequency ($\nu$), which matches the experimental results.
• Independence of Intensity: The intensity of light corresponds to the number of photons arriving per second. A higher intensity means more photons, which ejects more electrons (higher current), but the energy of each individual photon ($h\nu$) remains the same. Therefore, the maximum kinetic energy of the ejected electrons does not change with intensity.
• Existence of Threshold Frequency: For an electron to be emitted, its kinetic energy must be positive ($K_{\max} > 0$). This means $h\nu$ must be greater than $\phi_0$. The minimum frequency that can cause emission is when the photon has just enough energy to free the electron, so $h\nu_0 = \phi_0$. This gives the threshold frequency: $\nu_0 = \frac{\phi_0}{h}$ If the incident light's frequency $\nu$ is less than $\nu_0$, no electrons will be emitted, no matter how intense the light is.
• Instantaneous Process: Since the energy transfer happens in a single event (one photon to one electron), the process is instantaneous. There is no time delay for energy to build up.

Since $K_{\max} = eV_0$, we can rewrite Einstein's equation as: $eV_0 = h\nu - \phi_0$ or $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$ This equation predicts that a graph of stopping potential ($V_0$) versus frequency ($\nu$) is a straight line with a slope of $h/e$, which was experimentally verified by R. A. Millikan. His experiments confirmed Einstein's equation and provided an accurate value for Planck's constant.

Particle Nature of Light: The Photon

The photoelectric effect provided strong evidence that light, in its interaction with matter, behaves as if it is made of particles called photons. Einstein also proposed that these light quanta have momentum.

Properties of Photons:

• In interactions with matter, radiation acts like it's composed of particles called photons.
• Each photon has energy $E = h\nu$ and momentum $p = h\nu/c = h/\lambda$. Photons travel at the speed of light, $c$.
• All photons of a particular frequency $\nu$ have the same energy and momentum, regardless of the light's intensity. Higher intensity simply means more photons per second.
• Photons are electrically neutral and are not affected by electric or magnetic fields.
• In photon-particle collisions (like a photon hitting an electron), total energy and total momentum are conserved.

Given

• Frequency, $\nu = 6.0 \times 10^{14} \text{ Hz}$
• Power, $P = 2.0 \times 10^{-3} \text{ W}$
• Planck's constant, $h = 6.63 \times 10^{-34} \text{ J s}$

To Find

(a) Energy of a photon, $E$ (b) Number of photons per second, $N$

Formula

$E = h\nu$ $P = NE$

Solution

(a) Calculate the energy of a single photon $E = (6.63 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ Hz})$ $E = 3.98 \times 10^{-19} \text{ J}$ Answer for part (a) = $3.98 \times 10^{-19} \text{ J}$

(b) Calculate the number of photons emitted per second The total power $P$ is the number of photons per second $N$ multiplied by the energy of each photon $E$. $N = \frac{P}{E}$ $N = \frac{2.0 \times 10^{-3} \text{ W}}{3.98 \times 10^{-19} \text{ J}}$ $N = 5.0 \times 10^{15} \text{ photons per second}$ Answer for part (b) = $5.0 \times 10^{15}$ photons per second.

Given

• Work function, $\phi_0 = 2.14 \text{ eV} = 2.14 \times 1.6 \times 10^{-19} \text{ J}$
• Stopping potential, $V_0 = 0.60 \text{ V}$

To Find

(a) Threshold frequency, $\nu_0$ (b) Wavelength of incident light, $\lambda$

Formula

$\phi_0 = h\nu_0$ $eV_0 = h\nu - \phi_0$

Solution

(a) Calculate the threshold frequency The energy of a photon at the threshold frequency is equal to the work function. $\nu_0 = \frac{\phi_0}{h}$ $\nu_0 = \frac{2.14 \times 1.6 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}}$ $\nu_0 = 5.16 \times 10^{14} \text{ Hz}$ Answer for part (a) = $5.16 \times 10^{14} \text{ Hz}$. Light with a frequency below this value will not eject any photoelectrons from caesium.

(b) Calculate the wavelength of the incident light The maximum kinetic energy of the photoelectrons is given by $eV_0$. Using Einstein's photoelectric equation: $eV_0 = h\nu - \phi_0$ We can find the energy of the incident photons, $h\nu$: $h\nu = eV_0 + \phi_0$ $h\nu = (1.6 \times 10^{-19} \text{ C})(0.60 \text{ V}) + (2.14 \text{ eV})$ $h\nu = 0.60 \text{ eV} + 2.14 \text{ eV} = 2.74 \text{ eV}$ Now, convert this energy to Joules: $E = 2.74 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 4.384 \times 10^{-19} \text{ J}$ Since $E = h\nu = hc/\lambda$, we can solve for the wavelength $\lambda$: $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.63 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{4.384 \times 10^{-19} \text{ J}}$ $\lambda = 4.54 \times 10^{-7} \text{ m} = 454 \text{ nm}$ Answer for part (b) = $454 \text{ nm}$.

Wave Nature of Matter

Light exhibits a dual nature: it behaves like a wave in phenomena like diffraction and interference, but it behaves like a particle (photon) in phenomena like the photoelectric effect.

In 1924, Louis de Broglie proposed a bold hypothesis based on symmetry in nature. He reasoned that if radiation (energy) can have a particle-like nature, then matter (particles like electrons) should also have a wave-like nature.

He proposed that a moving particle with momentum $p$ has an associated wavelength $\lambda$, given by the de Broglie relation: $\lambda = \frac{h}{p} = \frac{h}{mv}$ where $m$ is the mass of the particle and $v$ is its speed. The wavelength $\lambda$ is called the de Broglie wavelength.

This equation beautifully connects a wave property (wavelength, $\lambda$) with a particle property (momentum, $p$).

For macroscopic objects that we see in everyday life, the de Broglie wavelength is incredibly small and impossible to measure. For example, a moving cricket ball has a wavelength so tiny that its wave nature is completely undetectable.

However, for sub-atomic particles like electrons, the wavelength can be significant and measurable, comparable to the spacing of atoms in a crystal. This wave nature of electrons was later experimentally confirmed and is a fundamental concept in quantum mechanics.

Given

(a) For the electron:

• Mass, $m = 9.11 \times 10^{-31} \text{ kg}$
• Speed, $v = 5.4 \times 10^6 \text{ m/s}$

(b) For the ball:

• Mass, $m' = 150 \text{ g} = 0.150 \text{ kg}$
• Speed, $v' = 30.0 \text{ m/s}$

To Find

(a) de Broglie wavelength of the electron, $\lambda$ (b) de Broglie wavelength of the ball, $\lambda'$

Formula

$\lambda = \frac{h}{p} = \frac{h}{mv}$

Solution

(a) Calculate the de Broglie wavelength of the electron First, find the momentum of the electron: $p = mv = (9.11 \times 10^{-31} \text{ kg}) \times (5.4 \times 10^6 \text{ m/s})$ $p = 4.92 \times 10^{-24} \text{ kg m/s}$ Now, calculate the wavelength: $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ J s}}{4.92 \times 10^{-24} \text{ kg m/s}}$ $\lambda = 1.35 \times 10^{-10} \text{ m} = 0.135 \text{ nm}$ Answer for part (a) = $0.135 \text{ nm}$. This is comparable to the wavelength of X-rays.

(b) Calculate the de Broglie wavelength of the ball First, find the momentum of the ball: $p' = m'v' = (0.150 \text{ kg}) \times (30.0 \text{ m/s})$ $p' = 4.50 \text{ kg m/s}$ Now, calculate the wavelength: $\lambda' = \frac{h}{p'} = \frac{6.63 \times 10^{-34} \text{ J s}}{4.50 \text{ kg m/s}}$ $\lambda' = 1.47 \times 10^{-34} \text{ m}$ Answer for part (b) = $1.47 \times 10^{-34} \text{ m}$. This wavelength is incredibly small, far beyond any possibility of measurement, which is why we don't observe the wave-like properties of macroscopic objects.