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Physics Chemistry Biology

Class 12th (Physics) Chapters
1. Electric Charges And Fields 2. Electrostatic Potential And Capacitance 3. Current Electricity
4. Moving Charges And Magnetism 5. Magnetism And Matter 6. Electromagnetic Induction
7. Alternating Current 8. Electromagnetic Waves 9. Ray Optics And Optical Instruments
10. Wave Optics 11. Dual Nature Of Radiation And Matter 12. Atoms
13. Nuclei 14. Semiconductor Electronics: Materials, Devices And Simple Circuits



11 Dual Nature Of Radiation And Matter



Introduction

By the late 19th century, the understanding of light was dominated by its wave nature, strongly supported by Maxwell's electromagnetic theory and experiments by Hertz demonstrating the generation and detection of electromagnetic waves. However, concurrent investigations into the conduction of electricity through gases at low pressures within discharge tubes led to discoveries that challenged the purely wave picture.

Significant milestones included the discovery of X-rays by Roentgen in 1895 and the electron by J.J. Thomson in 1897. Experiments with discharge tubes revealed that at very low pressures, an electric discharge occurred between the electrodes. A notable observation was a fluorescent glow on the glass wall opposite the cathode, attributed to rays emanating from the cathode, termed cathode rays. William Crookes first discovered these rays and suggested they were streams of fast-moving negatively charged particles.

J.J. Thomson confirmed this particle hypothesis experimentally. By subjecting cathode rays to perpendicular electric and magnetic fields, he was able to measure their velocity and the ratio of their charge to mass ($e/m$). The measured speeds were a fraction of the speed of light, and the $e/m$ value was found to be constant regardless of the cathode material or the gas in the tube. This universality suggested that these particles were fundamental constituents of matter.

Around the same period, it was observed that metals emitted negatively charged particles when exposed to ultraviolet light or when heated to high temperatures. The $e/m$ ratio for these particles was identical to that of cathode rays. These findings established that these particles, produced through different means, were fundamentally the same. J.J. Thomson named them electrons in 1897, proposing them as fundamental, universal components of matter. His groundbreaking work led to the Nobel Prize in Physics in 1906.

Later, in 1913, R.A. Millikan's famous oil-drop experiment precisely measured the charge of an electron, demonstrating that electric charge exists in discrete multiples of a fundamental unit, approximately $1.602 \times 10^{-19}$ C. Combining Millikan's charge value with Thomson's $e/m$ ratio allowed for the determination of the electron's mass.



Electron Emission

Metals are known to have a sea of "free" electrons that contribute to their electrical conductivity. These electrons are not bound to specific atoms but move freely within the metal lattice. However, they are generally confined within the metal because the surface of the metal exerts attractive forces (from the positively charged ions) that pull the electrons back. To escape the metal surface, an electron must overcome this attractive potential barrier.

The minimum energy required for an electron to escape from the surface of a metal is called the work function ($\phi_0$) of that metal. This property varies depending on the specific metal and the condition of its surface. The work function is commonly measured in units of electron volts (eV). One electron volt is defined as the kinetic energy gained by an electron when it is accelerated through an electric potential difference of 1 volt. $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$.

There are several ways to supply this minimum energy to the electrons to facilitate their emission from the metal surface:

(i) Thermionic Emission: When a metal is heated to a sufficiently high temperature, the thermal energy transferred to the free electrons can exceed the work function, allowing them to escape the surface.

(ii) Field Emission: Applying a very strong electric field (on the order of $10^8$ V/m) to the metal surface can pull electrons out. This is used in applications like spark plugs.

(iii) Photoelectric Emission: When light of a suitable frequency illuminates a metal surface, electrons are emitted. These emitted electrons are called photoelectrons, and the phenomenon is known as the photoelectric effect.



Photoelectric Effect

Hertz’s Observations

The first observation related to the photoelectric effect was made by Heinrich Hertz in 1887. While conducting experiments on generating and detecting electromagnetic waves using spark gaps, Hertz noticed that the sparks produced in his detector loop were more vigorous when the metal electrodes were illuminated by ultraviolet light from the source spark. This suggested that the ultraviolet light was somehow enhancing the electric discharge, implying that charged particles were being released from the metal surface by the action of light.


Hallwachs’ And Lenard’s Observations

Following Hertz's initial finding, Wilhelm Hallwachs and Philipp Lenard conducted detailed investigations into this effect. Lenard used an evacuated glass tube containing two metal electrodes. He observed that when ultraviolet radiation fell on one electrode (the emitter), a current flowed in the circuit, indicated by a galvanometer. The current stopped as soon as the radiation was removed. This confirmed that light was causing the emission of charged particles from the metal, which were then collected by the other electrode (the collector) due to the applied voltage.

Hallwachs further studied the effect using a charged zinc plate connected to an electroscope. He found that a negatively charged zinc plate lost its charge when illuminated by ultraviolet light. An uncharged zinc plate became positively charged under UV illumination, and a positively charged plate became even more positively charged. These observations strongly indicated that negatively charged particles were being emitted from the zinc surface under the influence of UV light.

After the electron was discovered in 1897, it became clear that the emitted negatively charged particles were electrons. Hallwachs and Lenard's experiments also revealed crucial characteristics of this emission: there was a minimum frequency of incident light (called the threshold frequency, $\nu_0$) below which no electrons were emitted, regardless of how intense the light was. This threshold frequency was found to be different for different metals. Metals like zinc, cadmium, and magnesium required ultraviolet light, while alkali metals like sodium and potassium were sensitive to visible light. Substances that exhibit this property are called photosensitive materials. The emitted electrons were termed photoelectrons, and the phenomenon itself was named the photoelectric effect.



Experimental Study Of Photoelectric Effect

A standard experimental setup to study the photoelectric effect involves an evacuated glass or quartz tube containing two metal electrodes: a photosensitive plate (emitter, C) and a collector plate (A) (

Photoelectric effect experimental setup
Figure 11.1). Light of controlled frequency and intensity is directed onto the emitter plate through a window. Electrons emitted from C are collected by A, creating a measurable electric current (photocurrent) in the external circuit. A battery provides a variable potential difference between C and A, and a commutator allows reversing the polarity so that A can be made positive (accelerating potential) or negative (retarding potential) relative to C. A voltmeter measures the potential difference, and a microammeter measures the photocurrent.

Using this apparatus, experiments were performed to study the variation of photocurrent with incident light intensity, frequency, the potential difference between electrodes, and the material of the emitter plate.


Effect Of Intensity Of Light On Photocurrent

With the frequency of the incident light and the potential difference between C and A kept constant (A at a positive potential to collect electrons), the intensity of the light falling on C is varied. It is observed that the photocurrent is directly proportional to the intensity of the incident light (

Photocurrent vs Light Intensity graph
Figure 11.2). Since photocurrent is a measure of the number of photoelectrons emitted per second, this experiment showed that the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.


Effect Of Potential On Photoelectric Current

Keeping the frequency and intensity of the incident light fixed, the potential of the collector plate A relative to the emitter plate C is varied. Initially, with A at a positive potential, increasing the positive voltage increases the photocurrent. At a certain positive potential, the current reaches a maximum value and does not increase further even if the voltage is raised. This is called the saturation current, indicating that all the photoelectrons emitted from C are being collected by A.

If the potential of A is made negative (retarding potential) with respect to C, the photocurrent decreases. Only electrons with sufficient kinetic energy can overcome the repulsive force from A and reach it. As the negative potential is increased, fewer electrons reach A, and the current drops. Eventually, at a specific minimum negative potential, the photocurrent becomes zero. This potential is called the stopping potential ($V_0$) or cut-off potential.

At the stopping potential $V_0$, even the most energetic photoelectrons with maximum kinetic energy ($K_{max}$) are just prevented from reaching the collector. Thus, the work done by the stopping potential on these electrons is equal to their maximum kinetic energy:

$$K_{max} = eV_0$$

where $e$ is the elementary charge of the electron.

Repeating this experiment with the same frequency but different intensities reveals that while the saturation current increases with increasing intensity (more electrons emitted), the stopping potential ($V_0$) remains unchanged (

Photocurrent vs Potential graph for different intensities
Figure 11.3). This means that for a given frequency, the maximum kinetic energy of the photoelectrons is independent of the intensity of the incident light.


Effect Of Frequency Of Incident Radiation On Stopping Potential

Now, keeping the intensity of incident light constant, the experiment is performed with different frequencies ($\nu$) of light. It is observed that for different frequencies (above a certain value), the stopping potential ($V_0$) required to stop the photocurrent is different (

Photocurrent vs Potential graph for different frequencies
Figure 11.4). Higher frequencies require a greater negative stopping potential, indicating that the maximum kinetic energy of photoelectrons increases with the frequency of incident light. The saturation current remains the same for different frequencies at the same intensity (since the number of emitted electrons is proportional to intensity).

A graph plotting the stopping potential ($V_0$) against the frequency ($\nu$) of the incident radiation for a given photosensitive material shows a straight line (

Stopping Potential vs Frequency graph
Figure 11.5). This graph has two key implications:

(i) The stopping potential, and thus the maximum kinetic energy of photoelectrons ($K_{max} = eV_0$), varies linearly with the frequency of incident radiation.

(ii) There is a minimum frequency, the threshold frequency ($\nu_0$), below which the stopping potential is zero (meaning no emission occurs, $K_{max}=0$), regardless of the intensity. This threshold frequency is characteristic of the material and corresponds to the frequency at which photoemission just begins.

These experiments also showed that if the frequency of incident light is above the threshold frequency, photoelectric emission is almost instantaneous, with no detectable time delay (less than $10^{-9}$ s), even for very low light intensities.

In summary, the key experimental observations of the photoelectric effect are:



Photoelectric Effect And Wave Theory Of Light

The wave picture of light, which successfully explained interference, diffraction, and polarisation, faced significant challenges in explaining the observed features of the photoelectric effect. According to the classical wave theory:

1. Energy in a wave is spread out continuously over the entire wavefront. Electrons at the metal surface should absorb energy continuously from the incident radiation.

2. A higher intensity wave means larger electric and magnetic field amplitudes, implying more energy is delivered to the surface per unit area per unit time. Therefore, the energy absorbed by each electron should increase with increasing light intensity. This would lead to the expectation that the maximum kinetic energy of emitted electrons should increase with intensity.

3. With continuous energy absorption, even light of low frequency should be able to cause electron emission if it's intense enough or shines for a long enough duration to allow electrons to accumulate sufficient energy to overcome the work function.

These predictions from the wave theory directly contradict the experimental observations:

Therefore, the classical wave theory fails to explain the fundamental characteristics of the photoelectric effect.



Einstein’s Photoelectric Equation: Energy Quantum Of Radiation

In 1905, Albert Einstein provided a revolutionary explanation for the photoelectric effect by proposing a new concept: that electromagnetic radiation (light) consists of discrete packets or quanta of energy, later called photons. The energy of each quantum is directly proportional to the frequency of the light, given by $E = h\nu$, where $h$ is Planck's constant (first introduced by Max Planck in the context of blackbody radiation).

Einstein proposed that in the photoelectric effect, an electron at the metal surface absorbs a single quantum of energy (photon) from the incident radiation. If the energy of this absorbed photon ($h\nu$) is greater than the minimum energy required to remove the electron from the metal surface (the work function, $\phi_0$), the electron is emitted. Any energy from the photon in excess of the work function is converted into the kinetic energy of the emitted electron. The most energetic electrons are those that absorb a photon at the surface and do not lose energy through collisions within the metal. Their maximum kinetic energy ($K_{max}$) is given by:

$$K_{max} = h\nu - \phi_0$$

This is known as Einstein's photoelectric equation. Electrons bound more tightly than the minimum work function (if any) or those that suffer collisions before escaping will have less than this maximum kinetic energy.

Einstein's equation elegantly explains all the experimental observations of the photoelectric effect:

Relating the stopping potential $V_0$ to $K_{max}$ ($K_{max} = eV_0$), Einstein's equation can be written as:

$$eV_0 = h\nu - \phi_0 \quad \text{for } \nu \ge \nu_0$$

or

$$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$$

This form directly matches the experimentally observed linear relationship between $V_0$ and $\nu$ (

Stopping Potential vs Frequency graph
Figure 11.5). The slope of the $V_0$ versus $\nu$ graph should be equal to $h/e$, and the intercept on the frequency axis gives the threshold frequency $\nu_0 = \phi_0/h$. Millikan's precise experiments measured this slope for various metals, and using the known value of the elementary charge $e$, he was able to determine Planck's constant $h$. The value he obtained was in excellent agreement with the value determined from unrelated phenomena like blackbody radiation. This provided strong experimental confirmation of Einstein's photoelectric equation and 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:

(a) The energy of a single photon ($E$) is given by $E = h\nu$, where $h$ is Planck's constant ($6.63 \times 10^{-34} \text{ J s}$) and $\nu$ is the frequency.

Given frequency $\nu = 6.0 \times 10^{14} \text{ Hz}$.

$$E = (6.63 \times 10^{-34} \text{ J s}) \times (6.0 \times 10^{14} \text{ s}^{-1})$$

$$E = 39.78 \times 10^{-20} \text{ J} = 3.978 \times 10^{-19} \text{ J}$$

The energy of a photon in the light beam is approximately $3.98 \times 10^{-19} \text{ J}$.

(b) The power emitted by the source ($P$) is the total energy emitted per second. Since the energy is carried by photons, the power is also the number of photons emitted per second ($N$) multiplied by the energy of each photon ($E$).

$P = N \times E$.

Given power $P = 2.0 \times 10^{-3} \text{ W}$.

$$N = \frac{P}{E} = \frac{2.0 \times 10^{-3} \text{ W}}{3.978 \times 10^{-19} \text{ J}}$$

$$N \approx 0.5028 \times 10^{16} \text{ photons/s}$$

$$N \approx 5.03 \times 10^{15} \text{ photons/s}$$

On average, approximately $5.03 \times 10^{15}$ photons are emitted by the source 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 $\phi_0 = 2.14 \text{ eV}$. We need to convert this to Joules for calculations involving Planck's constant in J s.

$\phi_0 = 2.14 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) = 3.428 \times 10^{-19} \text{ J}$.

(a) The threshold frequency ($\nu_0$) is the minimum frequency required for photoemission, which corresponds to the photon energy equaling the work function: $h\nu_0 = \phi_0$.

Using Planck's constant $h = 6.626 \times 10^{-34} \text{ J s}$:

$$\nu_0 = \frac{\phi_0}{h} = \frac{3.428 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}}$$

$$\nu_0 \approx 0.5173 \times 10^{15} \text{ Hz} = 5.173 \times 10^{14} \text{ Hz}$$

The threshold frequency for caesium is approximately $5.17 \times 10^{14} \text{ Hz}$.

(b) When the photocurrent is stopped by a potential $V_0 = 0.60 \text{ V}$, this is the stopping potential. The maximum kinetic energy of the photoelectrons is $K_{max} = eV_0$.

$K_{max} = (1.602 \times 10^{-19} \text{ C}) \times (0.60 \text{ V}) = 0.9612 \times 10^{-19} \text{ J}$.

Alternatively, we can express this in eV: $K_{max} = 0.60 \text{ eV}$.

Using Einstein's photoelectric equation: $K_{max} = h\nu - \phi_0$. We know $h\nu = hc/\lambda$, where $c$ is the speed of light ($3.00 \times 10^8 \text{ m/s}$) and $\lambda$ is the wavelength of incident light.

$$eV_0 = \frac{hc}{\lambda} - \phi_0$$

We need to find $\lambda$. Rearranging the equation:

$$\frac{hc}{\lambda} = eV_0 + \phi_0$$

$$\lambda = \frac{hc}{eV_0 + \phi_0}$$

Using energies in Joules:

$$\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m/s})}{(0.9612 \times 10^{-19} \text{ J}) + (3.428 \times 10^{-19} \text{ J})}$$

$$\lambda = \frac{19.878 \times 10^{-26} \text{ J m}}{4.3892 \times 10^{-19} \text{ J}}$$

$$\lambda \approx 4.530 \times 10^{-7} \text{ m}$$

Converting to nanometers ($1 \text{ nm} = 10^{-9} \text{ m}$):

$$\lambda \approx 453.0 \text{ nm}$$

Alternatively, using energies in eV:

$\phi_0 = 2.14 \text{ eV}$, $eV_0 = 0.60 \text{ eV}$. The energy of the incident photon $h\nu = eV_0 + \phi_0 = 0.60 \text{ eV} + 2.14 \text{ eV} = 2.74 \text{ eV}$.

Now convert the photon energy to Joules: $2.74 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) = 4.389 \times 10^{-19} \text{ J}$.

Then find wavelength using $E = hc/\lambda$:

$$\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m/s})}{4.389 \times 10^{-19} \text{ J}}$$

$$\lambda \approx \frac{19.878 \times 10^{-26}}{4.389 \times 10^{-19}} \text{ m} \approx 4.530 \times 10^{-7} \text{ m} = 453.0 \text{ nm}$$

The wavelength of the incident light is approximately 453 nm.



Particle Nature Of Light: The Photon

The successful explanation of the photoelectric effect by Einstein, based on the idea of discrete energy packets, provided strong evidence that light, when interacting with matter, behaves as if it is composed of particles. These energy quanta were later named photons.

Einstein further reasoned that if a light quantum possesses energy $E = h\nu$, it should also possess momentum. For a photon moving at the speed of light $c$, the momentum $p$ is related to energy by $E = pc$. Therefore, the momentum of a photon is $p = E/c = h\nu/c$. Since $\nu = c/\lambda$, the momentum can also be expressed as $p = h/\lambda$.

Possessing definite values of both energy and momentum is a characteristic property of particles. This solidified the idea that the light quantum could be associated with a particle-like entity. Further experimental evidence for the particle nature of light came from A.H. Compton's experiments in 1924 on the scattering of X-rays by electrons (Compton effect), where X-rays behaved as particles colliding with electrons, conserving both energy and momentum.

The key characteristics of photons, the particles of electromagnetic radiation, can be summarized as follows:



Wave Nature Of Matter

Having established that radiation (like light) exhibits a dual nature – behaving sometimes as a wave and sometimes as a particle – a profound question arose: Does matter, typically considered to be made of particles (electrons, protons, etc.), also possess wave-like properties?

In 1924, Louis Victor de Broglie proposed a bold hypothesis based on the idea of symmetry in nature. He reasoned that if radiation, which primarily exhibits wave phenomena, can also show particle characteristics, then matter, which is primarily seen as particles, should similarly exhibit wave characteristics under appropriate conditions. De Broglie proposed that a wave is associated with every moving material particle, and the wavelength ($\lambda$) of this "matter wave" is related to the particle's momentum ($p$) by the same relationship that applies to photons:

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

where $h$ is Planck's constant. For a particle of mass $m$ moving with velocity $v$, its momentum is $p = mv$. Thus, the de Broglie wavelength is given by:

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

This equation is known as the de Broglie relation and embodies the wave-particle duality of matter. The left side ($\lambda$) represents a wave attribute, while the right side (involving $m$ and $v$, or $p$) represents particle attributes. Planck's constant $h$ serves as the link between these two aspects.

De Broglie's hypothesis was revolutionary and needed experimental verification. The de Broglie wavelength is inversely proportional to the momentum ($mv$) of the particle. For macroscopic objects (large mass and/or velocity), the momentum is large, resulting in an extremely small de Broglie wavelength. For instance, a baseball or a bullet has a de Broglie wavelength far too small to be detected by any current experimental method. This is why we do not observe wave-like phenomena (like diffraction or interference) with everyday objects; they primarily exhibit particle behavior.

However, for microscopic particles like electrons, protons, neutrons, or atoms, the mass ($m$) is very small. Even with moderate velocities, their momentum ($p$) can be small enough to result in a de Broglie wavelength that is comparable to interatomic distances in crystals or spacings in diffraction gratings. In these cases, the wave nature of matter becomes significant and can be experimentally observed. The wave nature of electrons was first experimentally confirmed by Clinton Davisson and Lester Germer in 1927 through electron diffraction experiments, followed by George Paget Thomson (J.J. Thomson's son) shortly after. This experimental confirmation earned de Broglie the Nobel Prize in Physics in 1929.

The concept of matter waves is fundamental to quantum mechanics, where the behavior of particles is described by wave functions, and phenomena like electron diffraction through crystals are explained as constructive and destructive interference of these matter waves.

Example 11.3. What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4´106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?

Answer:

The de Broglie wavelength ($\lambda$) associated with a moving particle is given by $\lambda = h/p = h/(mv)$, where $h = 6.626 \times 10^{-34} \text{ J s}$ is Planck's constant, $m$ is the mass of the particle, and $v$ is its speed.

(a) For the electron:

Mass of electron $m_e = 9.109 \times 10^{-31} \text{ kg}$. Speed $v_e = 5.4 \times 10^6 \text{ m/s}$.

Momentum $p_e = m_e v_e = (9.109 \times 10^{-31} \text{ kg}) \times (5.4 \times 10^6 \text{ m/s}) = 4.919 \times 10^{-24} \text{ kg m/s}$.

de Broglie wavelength $\lambda_e = \frac{h}{p_e} = \frac{6.626 \times 10^{-34} \text{ J s}}{4.919 \times 10^{-24} \text{ kg m/s}}$.

$$\lambda_e \approx 1.347 \times 10^{-10} \text{ m}$$

Converting to nanometers: $1 \text{ nm} = 10^{-9} \text{ m}$, so $1.347 \times 10^{-10} \text{ m} = 0.1347 \times 10^{-9} \text{ m} = 0.1347 \text{ nm}$.

The de Broglie wavelength of the electron is approximately 0.135 nm. This wavelength is comparable to the spacing between atoms in crystals, which is why electron diffraction is observed.

(b) For the ball:

Mass of ball $m_b = 150 \text{ g} = 0.150 \text{ kg}$. Speed $v_b = 30.0 \text{ m/s}$.

Momentum $p_b = m_b v_b = (0.150 \text{ kg}) \times (30.0 \text{ m/s}) = 4.50 \text{ kg m/s}$.

de Broglie wavelength $\lambda_b = \frac{h}{p_b} = \frac{6.626 \times 10^{-34} \text{ J s}}{4.50 \text{ kg m/s}}$.

$$\lambda_b \approx 1.472 \times 10^{-34} \text{ m}$$

The de Broglie wavelength of the ball is approximately $1.47 \times 10^{-34}$ m. This wavelength is exceedingly small, many orders of magnitude smaller than the size of atomic nuclei. Therefore, wave properties are not observable for macroscopic objects like this ball.



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 $6 \times10^{14}\text{Hz}$ 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. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{–15} \text{ V s}$. Calculate the value of Planck’s constant.

Answer:

Question 11.6. The threshold frequency for a certain metal is $3.3 \times 10^{14} \text{ Hz}$. If light of frequency $8.2 \times 10^{14} \text{ Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Answer:

Question 11.7. 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 11.8. Light of frequency $7.21 \times 10^{14} \text{ Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^5 \text{ m/s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer:

Question 11.9. 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 11.10. 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 $1.0 \times 10^{–9} \text{ kg}$ drifting with a speed of $2.2 \text{ m/s}$?

Answer:

Question 11.11. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Answer: