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Chapter 11 Dual Nature Of Radiation And Matter
INTRODUCTION
By the end of the 19th century, the **wave nature of light** was well-established through Maxwell's theory of electromagnetism and Hertz's experiments on electromagnetic waves in 1887. Simultaneously, experiments on electric discharge through gases at low pressures led to significant discoveries about the fundamental constituents of matter.
Key findings during this era included the discovery of **X-rays** by Roentgen in 1895 and the **electron** by J. J. Thomson in 1897. Experiments showed that at very low gas pressure (around 0.001 mm Hg), applying an electric field caused a discharge. A glow appeared on the glass opposite the cathode, attributed to **cathode rays** emanating from the cathode. William Crookes first identified these rays in 1870, proposing in 1879 that they were streams of fast, negatively charged particles.
J. J. Thomson experimentally confirmed this hypothesis in 1897. By using perpendicular electric and magnetic fields, he measured the speed (0.1 to 0.2 times the speed of light) and the **specific charge ($e/m$)** of cathode ray particles. The value of $e/m$ ($1.76 \times 10^{11} \text{ C/kg}$) was found to be independent of the cathode material or the gas used, suggesting the **universal nature** of these particles.
Around the same time (1887), it was observed that metals emitted negatively charged particles when illuminated by ultraviolet light or when heated. These particles had the same $e/m$ ratio as cathode rays, confirming they were identical. J. J. Thomson named these particles **electrons** in 1897, proposing them as fundamental, universal constituents of matter. His work on electrons earned him the Nobel Prize in Physics in 1906.
In 1913, R. A. Millikan's oil-drop experiment precisely measured the electron's charge, finding it to be an integral multiple of $1.602 \times 10^{-19}$ C. This established the **quantisation of electric charge**. Using the measured values of $e$ and $e/m$, the electron's mass ($m$) could be determined.
ELECTRON EMISSION
Metals contain **free electrons** responsible for their conductivity. These electrons are held within the metal surface by the attractive forces of positive ions, preventing them from escaping under normal conditions. To escape, an electron needs to gain a certain minimum amount of energy to overcome this attractive pull.
This minimum energy required for an electron to escape from the metal surface is called the **work function** ($\phi_0$) of the metal. It is typically measured in **electron volts (eV)**.
An electron volt (eV) is the energy gained by an electron when accelerated through a potential difference of 1 volt. $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. This unit is common in atomic and nuclear physics.
The work function $\phi_0$ varies depending on the specific metal and its surface properties. For example, platinum has a high work function ($\phi_0 = 5.65 \text{ eV}$), while caesium has a low work function ($\phi_0 = 2.14 \text{ eV}$).
The energy needed for electron emission can be supplied through several physical processes:
- Thermionic emission: Heating the metal provides sufficient thermal energy to the free electrons for them to escape.
- Field emission: Applying a very strong external electric field ($\sim 10^8 \text{ V/m}$) can pull electrons out of the metal surface.
- Photoelectric emission: Illuminating the metal surface with light of suitable frequency causes electrons to be emitted. These emitted electrons are called **photoelectrons**.
PHOTOELECTRIC EFFECT
The emission of electrons from a metal surface when light falls on it is known as the **photoelectric effect**. This phenomenon was first observed by Heinrich Hertz in 1887.
Hertz’S Observations
During his electromagnetic wave experiments, Hertz observed that sparks produced were stronger when the emitter plate was illuminated by ultraviolet light. This suggested that light was somehow assisting the release of charged particles from the metal surface. Electrons near the surface absorb energy from the incident light, enabling them to overcome the binding forces and escape.
Hallwachs’ And Lenard’S Observations
Wilhelm Hallwachs and Philipp Lenard conducted more detailed investigations (1886-1902). Lenard observed that current flowed in a vacuum tube when ultraviolet radiation struck the emitter plate and stopped when the radiation ceased (Figure 11.1). This confirmed that light caused the emission of electrons, which formed a current.
Hallwachs' experiments with a charged zinc plate showed that UV light caused it to lose negative charge or gain positive charge, further indicating the emission of negative particles (electrons). After the electron's discovery, these were identified as photoelectrons.
A crucial observation by Hallwachs and Lenard was that photoemission only occurs if the incident light's frequency is greater than a certain minimum value, called the **threshold frequency** ($\nu_0$). This frequency is characteristic of the metal; below $\nu_0$, no emission happens, regardless of light intensity. Metals like zinc require UV, while alkali metals are sensitive to visible light.
EXPERIMENTAL STUDY OF PHOTOELECTRIC EFFECT
A standard experimental setup (Figure 11.1) is used to study the photoelectric effect, involving a photosensitive emitter (C), a collector (A), and an ac source with variable voltage/polarity, illuminated by monochromatic light. Photocurrent is measured with a microammeter.
Effect Of Intensity Of Light On Photocurrent
With fixed frequency and potential, increasing the intensity of incident light leads to a **linear increase in photoelectric current** (Figure 11.2). This means the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation.
Effect Of Potential On Photoelectric Current
With fixed frequency and intensity, varying the potential difference between the plates:
- As the collector A is made more positive (accelerating potential), the photocurrent increases until it reaches a maximum value called **saturation current**, where all emitted photoelectrons are collected.
- As A is made negative (retarding potential), the photocurrent decreases. At a specific minimum negative potential, the **stopping potential ($V_0$)**, the photocurrent becomes zero. This voltage stops even the most energetic photoelectrons. The maximum kinetic energy ($K_{max}$) of emitted electrons is related to $V_0$ by $\mathbf{K_{max} = eV_0}$ (Equation 11.1).
Figure 11.3 shows that increasing light intensity increases saturation current but **does not change the stopping potential** for a given frequency. $K_{max}$ is independent of intensity.
Effect Of Frequency Of Incident Radiation On Stopping Potential
With fixed intensity, varying the frequency ($\nu$) of incident light:
- Different frequencies result in different stopping potentials (Figure 11.4). Higher frequencies require a more negative $V_0$, indicating higher $K_{max}$ for photoelectrons.
Plotting $V_0$ versus $\nu$ yields a **straight line** for each metal (Figure 11.5). The slope is constant, but the intercept on the frequency axis varies, representing the **threshold frequency ($\nu_0$)** below which $V_0$ is zero (no emission). $K_{max}$ depends linearly on $\nu$ but not intensity.
Summary of experimental observations:
- Photocurrent $\propto$ intensity (above $\nu_0$).
- $V_0$ (and $K_{max}$) depends on $\nu$ and material, not intensity.
- Threshold frequency $\nu_0$ exists; no emission below it.
- Photoemission is practically **instantaneous** ($\sim 10^{-9}$ s), even for low intensity, if $\nu > \nu_0$.
PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT
The classical wave theory described light as a continuous electromagnetic wave. It predicted that energy absorption by electrons would be continuous and distributed over the wavefront.
Based on this wave picture, the following would be expected for photoelectric emission:
- $K_{max}$ of photoelectrons should increase with light intensity (higher amplitude means more energy absorbed). **Contradicts observation.**
- A threshold frequency should not exist; any frequency, given enough time and intensity, should supply enough energy. **Contradicts observation.**
- Energy absorption by individual electrons would take time, especially at low intensity. Emission should not be instantaneous. **Contradicts observation.**
The wave picture of light fundamentally failed to explain the key features of the photoelectric effect.
EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION
In 1905, Albert Einstein proposed a revolutionary picture of electromagnetic radiation to explain the photoelectric effect. He suggested that light energy is not continuous but consists of discrete packets or bundles of energy called **quanta** (later called **photons**).
Each photon has energy $\mathbf{E = h\nu}$, where $h$ is Planck's constant and $\nu$ is the light frequency. Photoelectric emission occurs when an electron absorbs a single photon. If the photon energy ($h\nu$) exceeds the minimum energy needed for the electron to escape the metal ($\phi_0$, work function), the electron is emitted with maximum kinetic energy.
$\mathbf{K_{max} = h\nu - \phi_0}$ (Equation 11.2)
This is **Einstein's photoelectric equation**, based on energy conservation. Energy of photon = Work function + Maximum kinetic energy of emitted electron.
This picture successfully explains the experimental observations:
- $K_{max}$ depends linearly on $\nu$ and is independent of intensity: The equation directly shows this. Emission results from a single photon absorption, so its energy ($h\nu$) determines $K_{max}$, not the number of photons (intensity).
- Threshold frequency ($\nu_0$) exists: For emission, $K_{max} \ge 0$, so $h\nu \ge \phi_0$. This implies a minimum frequency $\mathbf{\nu_0 = \phi_0/h}$ (Equation 11.3) is required.
- Photocurrent $\propto$ intensity: Higher intensity means more photons per second, leading to more electrons absorbing photons and being emitted, thus increasing photocurrent (for $\nu > \nu_0$).
- Instantaneous emission: The elementary process of photon absorption is instantaneous. If $h\nu \ge \phi_0$, emission happens immediately. Low intensity means fewer events, not delayed events.
Using $K_{max} = eV_0$ (Equation 11.1), the equation can be written as $\mathbf{eV_0 = h\nu - \phi_0}$, or $\mathbf{V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}}$ (Equation 11.4). This predicts a linear $V_0$ vs $\nu$ graph with slope $h/e$, independent of the material. Millikan's experiments (1906-1916) confirmed this linearity and determined $h$, matching Planck's value, solidifying the acceptance of the photon concept.
Example 11.1. Monochromatic light of frequency 6.0 ×1014 Hz is produced by a laser. The power emitted is 2.0 ×10–3 W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source?
Answer:
Given: Frequency $\nu = 6.0 \times 10^{14}$ Hz. Power $P = 2.0 \times 10^{-3}$ W. Planck's constant $h \approx 6.63 \times 10^{-34}$ J s.
(a) Energy of one photon $E = h\nu$.
$E = (6.63 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ Hz}) = 3.978 \times 10^{-19}$ J.
$E \approx 3.98 \times 10^{-19}$ J.
(b) Power is energy emitted per second. If $N$ photons are emitted per second, $P = N \times E$. So $N = P/E$.
$N = \frac{2.0 \times 10^{-3} \text{ W}}{3.978 \times 10^{-19} \text{ J}} \approx 0.5028 \times 10^{16} \text{ s}^{-1} \approx 5.0 \times 10^{15} \text{ photons/s}$.
The energy of a photon is $\approx 3.98 \times 10^{-19}$ J. The source emits $\approx 5.0 \times 10^{15}$ photons per second.
Example 11.2. The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.
Answer:
Given: Work function $\phi_0 = 2.14$ eV. Stopping potential $V_0 = 0.60$ V. $e \approx 1.602 \times 10^{-19}$ C. $h \approx 6.63 \times 10^{-34}$ J s. $1 \text{ eV} \approx 1.602 \times 10^{-19}$ J.
(a) Threshold frequency $\nu_0 = \phi_0 / h$. Convert $\phi_0$ to joules: $\phi_0 = 2.14 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) \approx 3.428 \times 10^{-19}$ J.
$\nu_0 = \frac{3.428 \times 10^{-19} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} \approx 0.517 \times 10^{15} \text{ Hz} = 5.17 \times 10^{14} \text{ Hz}$. The text gives $5.16 \times 10^{14}$ Hz based on a rounded value of $1.6 \times 10^{-19}$ for eV conversion.
The threshold frequency for caesium is $\approx 5.17 \times 10^{14}$ Hz.
(b) Using Einstein's equation $eV_0 = h\nu - \phi_0$. The energy of the incident photon is $h\nu = eV_0 + \phi_0$. Convert all to J: $eV_0 = (1.602 \times 10^{-19} \text{ C})(0.60 \text{ V}) = 0.9612 \times 10^{-19}$ J. $\phi_0 = 3.428 \times 10^{-19}$ J (from part a).
$h\nu = 0.9612 \times 10^{-19} \text{ J} + 3.428 \times 10^{-19} \text{ J} = 4.3892 \times 10^{-19}$ J.
Wavelength $\lambda = hc/(h\nu)$. $c = 3 \times 10^8$ m/s.
$\lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{4.3892 \times 10^{-19} \text{ J}} = \frac{19.89 \times 10^{-26}}{4.3892 \times 10^{-19}} \text{ m} \approx 4.53 \times 10^{-7} \text{ m} = 453 \text{ nm}$. The text gives 454 nm.
The wavelength of incident light is $\approx 453$ nm.
Example 11.3. The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour. (a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.) (b) From which of the photosensitive materials with work functions listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a), can you build a photoelectric device that operates with visible light?
Answer:
Given: $h = 6.63 \times 10^{-34}$ J s, $1 \text{ eV} = 1.6 \times 10^{-19}$ J. Speed of light $c \approx 3 \times 10^8$ m/s.
Energy of a photon $E = h\nu = hc/\lambda$. It's convenient to use $hc$ in eV nm unit for visible light. $hc = (6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s}) = 1.989 \times 10^{-25} \text{ J m}$. Convert to eV nm: $1 \text{ m} = 10^9 \text{ nm}$, $1 \text{ J} = 1/(1.6 \times 10^{-19}) \text{ eV}$.
$hc = (1.989 \times 10^{-25} \text{ J m}) \times \frac{1 \text{ eV}}{1.6 \times 10^{-19} \text{ J}} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = \frac{1.989 \times 10^{-25} \times 10^9}{1.6 \times 10^{-19}} \text{ eV nm} = \frac{1.989 \times 10^{-16}}{1.6 \times 10^{-19}} \text{ eV nm} = 1.243 \times 10^3 \text{ eV nm} = 1243 \text{ eV nm}$.
So, $E(\text{in eV}) = \frac{1243}{\lambda(\text{in nm})}$.
(a) Energies of photons:
- (i) Violet light: $\lambda_1 = 390$ nm. $E_1 = \frac{1243}{390} \text{ eV} \approx 3.187 \text{ eV}$. Text gives 3.19 eV.
- (ii) Yellow-green light: $\lambda_2 = 550$ nm. $E_2 = \frac{1243}{550} \text{ eV} \approx 2.2509 \text{ eV}$. Text gives 2.26 eV.
- (iii) Red light: $\lambda_3 = 760$ nm. $E_3 = \frac{1243}{760} \text{ eV} \approx 1.6355 \text{ eV}$. Text gives 1.64 eV.
Photon energies: (i) $\approx 3.19$ eV, (ii) $\approx 2.26$ eV, (iii) $\approx 1.64$ eV.
(b) Photosensitive materials for visible light: A photoelectric device works if the incident photon energy $E$ is greater than or equal to the work function $\phi_0$ of the metal ($E \ge \phi_0$). From Table 11.1 (provided in the text), work functions for some metals in eV are: Cs (2.14), K (2.30), Na (2.75), Ca (3.20), Mo (4.17), Pb (4.25), Al (4.28), Hg (4.49), Cu (4.65), Ag (4.70), Ni (5.15), Pt (5.65).
- Violet light ($E_1 \approx 3.19$ eV): Can cause emission from metals with $\phi_0 \le 3.19$ eV. From the list: Cs, K, Na, Ca.
- Yellow-green light ($E_2 \approx 2.26$ eV): Can cause emission from metals with $\phi_0 \le 2.26$ eV. From the list: Cs, K (borderline). Text says only Cs. Let's assume K is just above 2.26.
- Red light ($E_3 \approx 1.64$ eV): Can cause emission from metals with $\phi_0 \le 1.64$ eV. None of the listed metals have a work function this low.
To build a photoelectric device that operates with visible light, the material must have a work function less than or equal to the energy of the visible light photons incident on it. Based on the lowest energy visible light (red, $\sim 1.64$ eV), none of these metals will work for red light. But for yellow-green and violet light, metals like **Caesium (Cs)** ($\phi_0 = 2.14$ eV) will work (sensitive to yellow-green and violet), **Potassium (K)** ($\phi_0 = 2.30$ eV) will work (sensitive to violet), **Sodium (Na)** ($\phi_0 = 2.75$ eV) will work (sensitive to violet). **Calcium (Ca)** ($\phi_0 = 3.20$ eV) will work with violet only (borderline). Therefore, metals like **Cs, K, Na, and Ca** can be used to build devices that respond to at least some part of the visible spectrum.
WAVE NATURE OF MATTER
The previous discussion highlights the **dual nature of radiation** (light), exhibiting both wave-like (interference, diffraction, polarisation) and particle-like (photoelectric effect, Compton effect) properties depending on the phenomenon observed. The wave picture is useful for propagation (like focusing by a lens), while the particle picture is needed for interaction/absorption (like by retinal cells).
A natural question arises: If radiation (energy) has this dual nature, does **matter** (particles like electrons, protons, etc.) also exhibit wave-like properties? In 1924, French physicist Louis Victor de Broglie proposed the bold hypothesis that moving material particles should also possess wave characteristics under suitable conditions, based on the symmetry between matter and energy.
De Broglie proposed that the wavelength ($\lambda$) associated with a moving particle is related to its momentum ($p$) by:**
$\mathbf{\lambda = \frac{h}{p} = \frac{h}{mv}}$ (Equation 11.5)
where $h$ is Planck's constant, $m$ is the particle's mass, and $v$ is its speed. This is the **de Broglie relation**, and the associated wave is called a **matter wave** or **de Broglie wave**.
The de Broglie relation shows the duality: $\lambda$ (wave property) is related to $p=mv$ (particle property) by $h$. This relation also holds for photons ($\lambda_{photon} = c/\nu = h/(h\nu/c) = h/p_{photon}$).
From the formula, wavelength $\lambda$ is smaller for more massive particles (large $m$) or faster particles (large $v$). For macroscopic objects in everyday life (large $m$, significant $v$), $p=mv$ is very large, leading to an extremely small de Broglie wavelength ($\sim 10^{-34}$ m for a ball). This makes the wave properties of macroscopic objects unobservable in practice.
However, for sub-atomic particles like electrons (small $m$), even at relatively low speeds, the de Broglie wavelength can be significant and measurable, comparable to atomic dimensions. For an electron accelerated from rest through a potential $V$, its kinetic energy $K = eV$, momentum $p = \sqrt{2mK} = \sqrt{2meV}$. Its de Broglie wavelength is $\lambda = h/\sqrt{2meV}$. Substituting numerical values, this simplifies to $\mathbf{\lambda \approx \frac{1.227}{\sqrt{V}} \text{ nm}}$ when $V$ is in volts and $\lambda$ in nanometers (Equation 11.11). For $V=120$ V, $\lambda \approx 0.112$ nm, comparable to atomic spacing in crystals. This suggested that electron wave nature could be verified by crystal diffraction experiments.
De Broglie received the Nobel Prize in Physics in 1929 for his hypothesis.
Photocell
A **photocell** is a device that utilises the photoelectric effect to convert changes in light intensity into changes in electric current. It is sometimes called an "electric eye".
It typically consists of a photosensitive metal plate (cathode) and a collector (anode) inside an evacuated glass or quartz bulb. When light of suitable wavelength (frequency above the threshold) falls on the photosensitive cathode, photoelectrons are emitted. These electrons are attracted to the anode by an applied voltage, resulting in a measurable photocurrent.
Photocells have numerous applications:
- Light detection and measurement (e.g., in cameras).
- Control systems (e.g., automatic door openers, counting devices, burglar/fire alarms that respond to interrupted light beams).
- Sound reproduction in motion pictures.
- Scanning in television cameras.
- Detecting flaws in industrial production lines.
Probability Interpretation To Matter Waves
Understanding the precise physical meaning of a matter wave is challenging. Max Born provided a crucial interpretation: the **intensity** (square of the amplitude, $|A|^2$) of the matter wave at a point in space is proportional to the **probability density** of finding the particle at that point. Probability density means probability per unit volume.
So, $|A|^2 \Delta V$ is the probability of finding the particle in a small volume $\Delta V$ around that point. Where the wave intensity is high, the probability of finding the particle is high.
This interpretation connects the wave picture (amplitude) to the particle picture (probability of location). For a particle with a definite momentum, its de Broglie wave has a single wavelength and extends over all space. Born's rule implies the particle is not localized, meaning its position uncertainty is infinite, consistent with Heisenberg's uncertainty principle ($\Delta x \Delta p \approx \hbar$), where zero momentum uncertainty implies infinite position uncertainty.
More realistically, a particle is described by a **wave packet**, a superposition of waves localised in space ($\Delta x$ finite). A wave packet inherently has a spread of wavelengths, leading to a spread of momenta ($\Delta p$) via the de Broglie relation, consistent with the uncertainty principle ($\Delta x \Delta p \approx \hbar$).
Example 11.4. What is the de Broglie wavelength associated with (a) an electron moving with a speed of $5.4 \times 10^6$ m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?
Answer:
Using the de Broglie relation $\lambda = h/p = h/(mv)$, with Planck's constant $h \approx 6.63 \times 10^{-34}$ J s.
(a) For the electron: Mass $m_e = 9.11 \times 10^{-31}$ kg, speed $v_e = 5.4 \times 10^6$ m/s.
Momentum $p_e = m_e v_e = (9.11 \times 10^{-31} \text{ kg}) \times (5.4 \times 10^6 \text{ m/s}) = 4.9194 \times 10^{-24} \text{ kg m/s}$.
$\lambda_e = \frac{6.63 \times 10^{-34} \text{ J s}}{4.9194 \times 10^{-24} \text{ kg m/s}} \approx 1.3477 \times 10^{-10} \text{ m}$. Convert to nm: $1.3477 \times 10^{-10} \text{ m} = 0.13477 \times 10^{-9} \text{ m} = 0.13477 \text{ nm}$. The text gives 0.135 nm.
The de Broglie wavelength of the electron is $\approx 0.135$ nm.
(b) For the ball: Mass $m_{ball} = 150 \text{ g} = 0.150 \text{ kg}$, speed $v_{ball} = 30.0 \text{ m/s}$.
Momentum $p_{ball} = m_{ball} v_{ball} = (0.150 \text{ kg}) \times (30.0 \text{ m/s}) = 4.50 \text{ kg m/s}$.
$\lambda_{ball} = \frac{6.63 \times 10^{-34} \text{ J s}}{4.50 \text{ kg m/s}} = 1.4733... \times 10^{-34} \text{ m}$. The text gives $1.47 \times 10^{-34}$ m.
The de Broglie wavelength of the ball is $\approx 1.47 \times 10^{-34}$ m.
Comparing the wavelengths: The electron's wavelength is in the range of X-rays or atomic spacing, hence wave effects are observable. The ball's wavelength is extremely small, far below measurable scales, so wave effects are not observed for macroscopic objects.
Example 11.5. An electron, an a-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Answer:
De Broglie wavelength $\lambda = h/p$. Kinetic energy $K = p^2/(2m)$, so momentum $p = \sqrt{2mK}$.
Thus, $\lambda = \frac{h}{\sqrt{2mK}}$.
For the same kinetic energy $K$, $\lambda$ is inversely proportional to the square root of the mass $m$ ($\lambda \propto 1/\sqrt{m}$). The particle with the largest mass will have the shortest de Broglie wavelength.
Compare masses: $m_e \approx 9.11 \times 10^{-31}$ kg. $m_p \approx 1.67 \times 10^{-27}$ kg $\approx 1836 m_e$. An $\alpha$-particle is a Helium nucleus ($^4_2\text{He}$), composed of 2 protons and 2 neutrons. $m_\alpha \approx 2m_p + 2m_n$. Since $m_n \approx m_p$, $m_\alpha \approx 4 m_p \approx 4 \times 1.67 \times 10^{-27}$ kg $\approx 6.68 \times 10^{-27}$ kg. (The text gives $m_\alpha \approx 4 m_p$, which is fine for comparison).
Masses in increasing order: $m_e < m_p < m_\alpha$.
De Broglie wavelengths in increasing order (since $\lambda \propto 1/\sqrt{m}$): $\lambda_\alpha < \lambda_p < \lambda_e$.
The particle with the shortest de Broglie wavelength is the **$\alpha$-particle** (due to its largest mass).
Example 11.6. A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10–4. Calculate the particle’s mass and identify the particle.
Answer:
Let $v_p$ and $v_e$ be the speeds of the particle and electron, and $m_p$ and $m_e$ be their masses. Given $v_p = 3v_e$. Let $\lambda_p$ and $\lambda_e$ be their de Broglie wavelengths. Given $\lambda_p / \lambda_e = 1.813 \times 10^{-4}$.
Using the de Broglie relation $\lambda = h/(mv)$:
$\lambda_p = \frac{h}{m_p v_p}$ and $\lambda_e = \frac{h}{m_e v_e}$.
Take the ratio of the wavelengths: $\frac{\lambda_p}{\lambda_e} = \frac{h/(m_p v_p)}{h/(m_e v_e)} = \frac{m_e v_e}{m_p v_p}$.
Substitute the given relations: $1.813 \times 10^{-4} = \frac{m_e v_e}{m_p (3v_e)} = \frac{1}{3} \frac{m_e}{m_p}$.
Solve for $m_p$: $m_p = \frac{m_e}{3 \times 1.813 \times 10^{-4}} = \frac{m_e}{5.439 \times 10^{-4}}$.
$m_p \approx 1838.6 \times m_e$.
Using $m_e \approx 9.11 \times 10^{-31}$ kg:
$m_p \approx 1838.6 \times (9.11 \times 10^{-31} \text{ kg}) \approx 1.675 \times 10^{-27} \text{ kg}$.
This mass is very close to the mass of a proton ($m_p \approx 1.672 \times 10^{-27}$ kg) or a neutron ($m_n \approx 1.675 \times 10^{-27}$ kg). The particle is likely a **proton** or a **neutron**.
Example 11.7. What is the de Broglie wavelength associated with an electron, accelerated through a potential differnece of 100 volts?
Answer:
Given: Accelerating potential $V = 100$ V. For an electron accelerated through a potential difference $V$, its kinetic energy is $K = eV$. The de Broglie wavelength is $\lambda = h/\sqrt{2m_e K} = h/\sqrt{2m_e eV}$.
Using the simplified formula $\lambda = \frac{1.227}{\sqrt{V}}$ nm (Equation 11.11), valid for electrons:
$\lambda = \frac{1.227}{\sqrt{100}} \text{ nm} = \frac{1.227}{10} \text{ nm} = 0.1227 \text{ nm}$.
The de Broglie wavelength associated with the electron is approximately 0.123 nm. This is comparable to the wavelength of X-rays and the spacing between atoms in crystals.
DAVISSON AND GERMER EXPERIMENT
The wave nature of electrons, predicted by de Broglie, was experimentally verified by C.J. Davisson and L.H. Germer in 1927, and independently by G.P. Thomson in 1928, through **electron diffraction** experiments using crystals. Davisson and Thomson shared the Nobel Prize in 1937 for this discovery.
The Davisson-Germer experiment used an **electron gun** to produce a collimated beam of electrons accelerated to a controlled velocity by a variable voltage. This electron beam was directed to strike a nickel crystal. Electrons were scattered by the crystal's atoms in various directions.
An **electron detector** measured the intensity of the scattered electron beam at different scattering angles. The apparatus was in an evacuated chamber.
Davisson and Germer measured the scattered electron intensity for accelerating voltages from 44 V to 68 V. They observed a strong peak in intensity at an accelerating voltage of **54 V** for a scattering angle of **50°**. This peak indicated constructive interference of the electron waves scattered from the ordered arrangement of atoms in the nickel crystal, similar to how X-rays are diffracted by crystals.
From the diffraction pattern, they calculated the wavelength of the electron waves to be **0.165 nm**.
They then calculated the de Broglie wavelength for an electron accelerated through 54 V using the formula $\lambda = 1.227/\sqrt{V}$ nm:
$\lambda = \frac{1.227}{\sqrt{54}} \text{ nm} \approx 0.167 \text{ nm}$.
The excellent agreement between the experimentally determined wavelength (0.165 nm) and the theoretically predicted de Broglie wavelength (0.167 nm) provided strong confirmation of de Broglie's hypothesis and the wave nature of electrons.
Subsequent experiments, like electron double-slit experiments and interference with molecules, further confirmed the wave nature of matter. The de Broglie hypothesis is a cornerstone of quantum mechanics and has led to applications like the electron microscope, which uses the wave properties of electrons for high-resolution imaging.
SUMMARY
This chapter explores the **dual nature of radiation (light)** and proposes the **wave nature of matter**. It starts with the photoelectric effect, which provides strong evidence for the particle nature of light.
- **Work function ($\phi_0$)** is the minimum energy required for an electron to escape a metal surface. Emission can be by heat (thermionic), electric field (field emission), or light (photoelectric).
- **Photoelectric effect** is electron emission from metals illuminated by light of suitable frequency. Key features: photocurrent $\propto$ intensity (above $\nu_0$), $K_{max}$ (or stopping potential $V_0$) depends on $\nu$ and material but not intensity, threshold frequency $\nu_0$ exists (no emission below it), emission is instantaneous. $K_{max} = eV_0$.
- Classical wave theory failed to explain photoelectric effect features.
- **Einstein's photoelectric equation**: $K_{max} = h\nu - \phi_0$. Light energy is in discrete **photons** ($E=h\nu$). One photon-one electron interaction. Explains all observations. $V_0 = (h/e)\nu - \phi_0/e$. Millikan's experiments verified this and $h$.
- **Particle nature of light (photon):** Photon has energy $h\nu$ and momentum $p=h\nu/c = h/\lambda$. Confirmed by Compton effect.
- **Wave nature of matter:** De Broglie hypothesis: Moving particles also have wave properties. **De Broglie wavelength** $\lambda = h/p = h/(mv)$. Duality in $\lambda$ and $p$. Measurable for sub-atomic particles, negligible for macroscopic objects. $\lambda_{electron} = 1.227/\sqrt{V}$ nm.
- **Davisson-Germer experiment** (and G.P. Thomson): Experimental verification of electron wave nature through crystal diffraction, matching de Broglie's prediction.
- Matter waves incorporate Heisenberg's uncertainty principle and support Bohr's quantisation condition.
- **Photocell:** Technological application of photoelectric effect, converting light intensity to current changes.
POINTS TO PONDER
Reflections on concepts:
- Free electrons in metals are confined within the surface; $\phi_0$ is the minimum escape energy.
- Electron energies in metals have a distribution; $\phi_0$ corresponds to the least energy needed by the most energetic electrons.
- Photoelectric observations imply energy absorption in discrete units $h\nu$, suggesting quantisation of light energy, not necessarily that light *is* particles at all times and places.
- Stopping potential observations ($V_0$ vs intensity and frequency) are crucial evidence distinguishing wave vs photon pictures.
- De Broglie wavelength $\lambda$ is physically significant; phase velocity $v_p = \nu\lambda$ is not. Group velocity equals particle velocity.
Table of relevant physical quantities:
| Physical Quantity | Symbol | Dimensions | Unit | Remarks |
|---|---|---|---|---|
| Planck’s constant | $h$ | [ML$^2$T$^{–1}$] | J s | $E = h\nu$ |
| Stopping potential | $V_0$ | [ML$^2$T$^{–3}$A$^{–1}$] | V | $eV_0 = K_{max}$ |
| Work function | $\phi_0$ | [ML$^2$T$^{–2}$] | J; eV | $K_{max} = h\nu – \phi_0$ |
| Threshold frequency | $\nu_0$ | [T$^{–1}$] | Hz | $\nu_0 = \phi_0/h$ |
| de Broglie wavelength | $\lambda$ | [L] | m | $\lambda = h/p$ |
Exercises
Question 11.1. Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
Question 11.2. The work function of caesium metal is 2.14 eV. When light of frequency $6 \times 10^{14}$Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) Stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Answer:
Question 11.3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Question 11.4. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer:
Question 11.5. The energy flux of sunlight reaching the surface of the earth is $1.388 \times 10^3 \text{ W/m}^2$. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
Question 11.6. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{–15}$ V s. Calculate the value of Planck’s constant.
Answer:
Question 11.7. A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Answer:
Question 11.8. The threshold frequency for a certain metal is $3.3 \times 10^{14}$ Hz. If light of frequency $8.2 \times 10^{14}$ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Question 11.9. The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:
Question 11.10. Light of frequency $7.21 \times 10^{14}$ Hz is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^5$ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Question 11.11. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer:
Question 11.12. Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Question 11.13. What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Question 11.14. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Answer:
Question 11.15. What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass $1.0 \times 10^{–9}$ kg drifting with a speed of 2.2 m/s?
Answer:
Question 11.16. An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:
Question 11.17. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{–10}$m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Answer:
Question 11.18. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer:
Question 11.19. What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
ADDITIONAL EXERCISES
Question 11.20. (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be $1.76 \times 10^{11} \text{ C kg}^{–1}$.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?
Answer:
Question 11.21. (a) A monoenergetic electron beam with electron speed of $5.20 \times 10^6 \text{ m s}^{–1}$ is subject to a magnetic field of $1.30 \times 10^{–4}$ T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals $1.76 \times 10^{11}\text{C kg}^{–1}$.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Answer:
Question 11.22. An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~$10^{–2}$ mm of Hg). A magnetic field of $2.83 \times 10^{–4}$ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:
Question 11.23. (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Answer:
Question 11.24. In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two $\gamma$-rays of equal energy. What is the wavelength associated with each $\gamma$-ray? (1BeV = $10^9$ eV)
Answer:
Question 11.25. Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~$10^{–10} \text{ W m}^{–2}$). Take the area of the pupil to be about 0.4 cm$^2$, and the average frequency of white light to be about $6 \times 10^{14}$ Hz.
Answer:
Question 11.26. Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~$10^5 \text{ W m}^{–2}$) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer:
Question 11.27. Monochromatic radiation of wavelength 640.2 nm (1nm = $10^{–9}$ m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:
Question 11.28. A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
$\lambda_1 = 3650 Å, \lambda_2= 4047 Å, \lambda_3= 4358 Å, \lambda_4= 5461 Å, \lambda_5= 6907 Å$
The stopping voltages, respectively, were measured to be:
$V_{01} = 1.28 \text{ V}, V_{02} = 0.95 \text{ V}, V_{03} = 0.74 \text{ V}, V_{04} = 0.16 \text{ V}, V_{05} = 0 \text{ V}$
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be $1.6 \times 10^{–19}$ C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Answer:
Question 11.29. The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?
Answer:
Question 11.30. Light of intensity $10^{–5} \text{ W m}^{–2}$ falls on a sodium photo-cell of surface area 2 cm$^2$. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
Question 11.31. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) ($m_e=9.11 \times 10^{–31}$ kg).
Answer:
Question 11.32. (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. ($m_n = 1.675 \times 10^{–27}$ kg)
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:
Question 11.33. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Answer:
Question 11.34. The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of $10^{–15}$ m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Answer:
Question 11.35. Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
Answer:
Question 11.36. Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{–10}$ m.
[Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distintguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Answer:
Question 11.37. Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:
$E = h\nu$, $p = \frac{h}{\lambda}$
But while the value of $\lambda$ is physically significant, the value of $\nu$ (and therefore, the value of the phase speed $\nu \lambda$) has no physical significance. Why?
Answer: