Class 12 Physics Chapter 11 Dual Nature of Radiation and matter

Key Concepts: Energy Units

b) Photons
[Solution Description]
In the context of the photoelectric effect, energy absorption takes place in discrete packets known as photons. This follows from Planck’s theory and relates to the quantized nature of light.

Key Concepts: Energy Distribution and Emission Mechanisms

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion considers two types of electron emissions: photoelectric emission and thermionic emission. Photoelectric emission occurs when electrons absorb energy from incident photons, requiring a specific quantum of energy

$$\text{work function or } f_0$$,

whereas thermionic emission involves increasing thermal kinetic energy to overcome the work function.

For the assertion:

– Electrons with higher kinetic energy indeed have a greater chance to be emitted via photoelectric emission because they need less additional energy from photons, making them suitable for photoelectric processes over thermionic ones.

For the reason:

– The photoelectric effect is characterized by the absorption of discrete quantum packets of energy (photons) where each photon has an energy given by $$E = h\nu$$, where $$h$$ is Planck’s constant and $$\nu$$ is frequency.

Thus, both statements are true, and the reason correctly explains why electrons with higher kinetic energy are likelier to be involved in photoelectric emission: due to their ability to utilize the energy quanta efficiently provided by photons.

Hence, the correct answer is option (a).

Key Concepts: Energy Unit

b) Electron volt
[Solution Description] In thermionic emission, the work function is usually expressed in electron volts (eV). An electron volt is a unit of energy equal to approximately $$1.602 \times 10^{-19}$$ Joules. It is convenient to use eV because this unit closely matches the typical magnitudes of energies involved in electronic transitions in solids like metals.

Key Concepts: Condition Analysis

c) High electric field
[Solution Description] Field emission requires a strong electric field that enables electrons to overcome the potential barrier at the metal surface without the need for high temperatures or light exposure. The electric field must be around $$10^8 \, \text{V/m}$$ for field emission to take place.

Key Concepts: Photon Interaction

b) Photons can be absorbed or create new photons.

[Solution Description]

Photons interact with matter in specific ways due to their wave-particle duality. They can either be absorbed or cause the emission of new photons through processes such as the photoelectric effect or Compton scattering. Photons do not gradually lose energy but rather transfer it completely during interactions.

key concept: Complex Problem Solving

a) $$1.23 \times 10^{-10} \, \text{m}$$
[Solution Description] To find the de Broglie wavelength, we need to first determine the velocity of the electron when it is accelerated by the given voltage. The kinetic energy acquired by the electron is equal to the work done on it by the electric field, which is given by $$ eV $$, where $$ e = 1.6 \times 10^{-19} $$ C and $$ V = 100 $$ V. This gives the kinetic energy, $$ KE = eV = 1.6 \times 10^{-19} \times 100 $$.

Next, equate this kinetic energy with the expression for kinetic energy in terms of velocity:
$$
KE = \frac{1}{2}mv^2
$$

Substitute the values:
$$
1.6 \times 10^{-17} = \frac{1}{2} \times 9.11 \times 10^{-31} \times v^2
$$

Solving for $$ v $$:
$$
v^2 = \frac{2 \times 1.6 \times 10^{-17}}{9.11 \times 10^{-31}}
$$

$$
v = \sqrt{3.51 \times 10^{14}}
$$

$$
v \approx 5.93 \times 10^7 \, \text{m/s}
$$

Now use de Broglie’s formula for wavelength:
$$
\lambda = \frac{h}{mv}
$$

Substituting the known values:
$$
\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 5.93 \times 10^7}
$$

$$
\lambda \approx 1.23 \times 10^{-10} \, \text{m}
$$

Key Concepts: Einstein’s Equation

a) 2 eV

[Solution Description]

According to Einstein’s photoelectric equation:

$$K_{max} = E – \phi_0$$

Given that $$E = 5 \, \text{eV}$$ and $$\phi_0 = 3 \, \text{eV}$$, substituting these values gives:

$$K_{max} = 5 \, \text{eV} – 3 \, \text{eV} = 2 \, \text{eV}$$

Thus, the maximum kinetic energy is $$2 \, \text{eV}$$.

Photon Energy

b) $$3.315 \times 10^{-19} \, \text{J}$$
[Solution Description] The energy of a photon can be calculated using the formula $$E = h\nu$$. Let’s substitute the given values into the formula:

$$
E = (6.63 \times 10^{-34} \text{ Js}) \times (5 \times 10^{14} \text{ Hz})
$$

Calculating this, we get:

$$
E = 3.315 \times 10^{-19} \text{ J}
$$

Therefore, the energy of the photon is $$3.315 \times 10^{-19}$$ Joules.

key concept: Advanced Calculation, Theoretical Implications

a) $$\lambda = 2.21 \times 10^{-10} \text{ m}$$
[Solution Description]
The de Broglie wavelength $$\lambda$$ is given by the formula:
$$
\lambda = \frac{h}{p}
$$
where

$$h = 6.626 \times 10^{-34} \, \text{Js}$$
is Planck’s constant and $$p$$ is the momentum of the particle.

Substituting the given values,
$$
\lambda = \frac{6.626 \times 10^{-34}}{3 \times 10^{-24}}
$$
Simplifying the calculation,
$$
\lambda = 2.2087 \times 10^{-10} \text{ m}
$$

This result signifies that electrons, though particles, exhibit significant wave-like properties at the subatomic level, which validates the dual nature postulated by de Broglie.

Key Concepts: Stopping Potential and Frequency

c) Stopping potential increases.

[Solution Description] The stopping potential $$(V_0)$$ is related to the maximum kinetic energy of photoelectrons by $$K_{max} = eV_0$$, where $$e$$ is the electron charge. According to the Einstein photoelectric equation, $$K_{max} = h\nu – \phi$$, where $$h$$ is Planck’s constant, $$\nu$$ is the frequency, and $$\phi$$ is the work function. As frequency $$\nu$$ increases, $$K_{max}$$ also increases (assuming $$\phi$$ remains constant). Thus, $$V_0$$ must increase to maintain the equality $$K_{max} = eV_0$$. Therefore, when the frequency of the incident light increases, the stopping potential increases.

Key Concepts: Problem-Solving

b) $$2.43 \times 10^{-10} \text{ m}$$

[Solution Description]

The de Broglie wavelength $$\lambda$$ is calculated using the formula

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

where $$v = 3 \times 10^6 \text{ m/s}$$, $$m = 9.11 \times 10^{-31} \text {kg}$$, and $$h = 6.626 \times 10^{-34} \text{ Js}$$. So,

$$\lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(3 \times 10^6)}$$

$$= \frac{6.626 \times 10^{-34}}{2.733 \times 10^{-24}}$$

$$\approx 2.43 \times 10^{-10} \text{ m}$$

Key Concepts: Observation Recall

c) It stops
[Solution Description] In Philipp Lenard’s experiments on the photoelectric effect, it was observed that when ultraviolet radiations were stopped, the current flow also stopped. This is because the electrons emitted due to UV radiation ceased to be ejected, halting the current.

Key Concepts: Quantum Theory, Threshold Frequency

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the photoelectric effect supports the quantum theory of light by demonstrating that light energy is quantized. This is shown through the fact that electrons are ejected only when the light frequency reaches or exceeds a certain value, known as the threshold frequency, which aligns with the idea of quantized packets of energy (photons).

The reason given explains that if the frequency of the incident light is below this threshold frequency, no electrons will be emitted, no matter how intense the light is. This reason is true and directly supports the assertion because it highlights the necessity of quantized energy packets to exceed the work function for electron emission.

According to Einstein’s photoelectric equation $$K_{\text{max}} = h\nu – \phi_0$$, where $$h$$ is Planck’s constant, $$\nu$$ is frequency, and $$\phi_0$$ is the work function, an electron will be emitted only if $$h\nu > \phi_0$$. Hence, both the assertion and reason are true, and the reason correctly explains the assertion in terms of quantum theory.

key concept: Conservation Laws, Real-World Application

c) 2.5 eV
[Solution Description]
According to the conservation of energy, the energy of the incoming photon minus the work function equals the kinetic energy of the emitted electron:
$$
E_{\text{photon}} = \phi_0 + E_{\text{kinetic}}
$$
Given that $$E_{\text{photon}} = 5 \text{ eV}$$ and $$E_{\text{kinetic}} = 2.5 \text{ eV}$$,
$$
5 = \phi_0 + 2.5
$$
Solve for the work function $$\phi_0$$:
$$
\phi_0 = 5 – 2.5 = 2.5 \text{ eV}
$$

key concept: Threshold Frequency

b) No threshold frequency should exist according to wave theory.
[Solution Description] According to wave theory, energy absorption by electrons is continuous and does not involve a threshold frequency. Therefore, there should be no threshold frequency for photoelectric emission.

Key Concepts: Comparative Analysis

b) Field emission
[Solution Description] Field emission involves releasing electrons by applying a strong electric field. This is different from thermionic emission, which relies on heating, and photoelectric emission, which depends on light.

key concept: De Broglie Wavelength

a) $$3.64 \times 10^{-10} \, \text{m}$$

[Solution Description]
The de Broglie wavelength $$\lambda$$ can be calculated using the formula $$\lambda = \frac{h}{mv}$$, where $$m$$ is the mass and $$v$$ is the velocity.

Step 1: Substitute the given values into the formula:
$$
\lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{9.11 \times 10^{-31} \, \text{kg} \times 2.0 \times 10^{6} \, \text{m/s}}
$$

Step 2: Perform the calculation:
$$
\lambda = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} \approx 3.64 \times 10^{-10} \, \text{m}
$$

Therefore, the de Broglie wavelength is approximately

$$3.64 \times 10^{-10} \, \text{m}$$
.

key concept: De Broglie Hypothesis Application, Experimental Verification

c) $$3.17 \times 10^{-10} \, \text{m}$$
[Solution Description] The kinetic energy $$E_k$$ of an electron can be related to its momentum by the equation $$E_k = \frac{p^2}{2m_e}$$. Solving for momentum $$p$$, we have:

$$
p = \sqrt{2m_eE_k}
$$

Since $$E_k = 150 \, \text{eV}$$ must be converted to joules:

$$
E_k = 150 \times 1.602 \times 10^{-19} \, \text{J} = 2.403 \times 10^{-17} \, \text{J}
$$

Thus, the momentum $$p$$ becomes:

$$
p = \sqrt{2 \times 9.11 \times 10^{-31} \, \text{kg} \times 2.403 \times 10^{-17} \, \text{J}}
= \sqrt{4.379 \times 10^{-47}}
= 2.092 \times 10^{-23} \, \text{kg}\cdot\text{m/s}
$$

Using De Broglie’s relation $$ \lambda = \frac{h}{p} $$, calculate $$\lambda$$:

$$
\lambda = \frac{6.626 \times 10^{-34} \, \text{J}\cdot\text{s}}{2.092 \times 10^{-23} \, \text{kg}\cdot\text{m/s}}
= 3.17 \times 10^{-10} \, \text{m}
$$

Key Concepts: Stopping Potential

d) It increases as the frequency of light increases

[Solution Description] Stopping potential is the minimum negative potential needed to stop the most energetic photoelectrons from reaching the anode. According to Einstein’s photoelectric equation: $$eV_0 = hf – \phi$$, where $$e$$ is the charge of electron, $$V_0$$ is the stopping potential, $$h$$ is Planck’s constant, $$f$$ is the frequency of the incident light, and $$\phi$$ is the work function. As frequency increases, so does the stopping potential.

Key Concepts: Basic Setup

b) Emitter plate
[Solution Description] The photosensitive plate, also known as the emitter, is made of a material that releases electrons when it absorbs energy from incident light. This process occurs during the photoelectric effect experiment.

Key Concepts: Basic Discovery

d) Heinrich Hertz
[Solution Description] Heinrich Hertz was a German physicist who is most well-known for his discovery of electromagnetic waves. In 1887, during his experiments with electric sparks, he discovered the phenomenon that later became known as the photoelectric effect, where ultraviolet light enhanced the spark length.

key concept: Photon Properties

b) Photons carry energy and momentum.
[Solution Description] Photons are particles that have no rest mass, but they do carry energy and momentum. They behave as both waves and particles, which is a fundamental concept in quantum mechanics.

key concept: Energy Quanta

a) $$h\nu$$
[Solution Description] Einstein proposed that light consists of quanta or photons, with each photon having an energy given by the formula $$E = h\nu$$, where $$h$$ is Planck’s constant and $$\nu$$ is the frequency of light.

Key Concepts: Intensity Effect

c) Saturation current increases.
[Solution Description] The saturation current is determined by the number of photoelectrons emitted, which depends on the intensity of the incident light. Higher intensity means more photons are available to release more electrons, thus increasing the saturation current.

key concept: Frequency Effect

c) It increases
[Solution Description]
The stopping potential $$V_0$$ is directly proportional to the frequency $$n$$ of the incident radiation. As the frequency increases, the energy of the photoelectrons ejected also increases. This results in a higher stopping potential required to stop them.

Key Concepts: Photon Characteristics

b) Frequency of light
[Solution Description] The energy of a photon is dependent on the frequency of the light. According to the relation $$E = hf$$, where $$h$$ is Planck’s constant and $$f$$ is the frequency.

key concept: Basic Concept

b) The number of electrons emitted per second
[Solution Description] According to Einstein’s postulates, the intensity of radiation affects the number of electrons emitted, which is reflected in the photoelectric current, for frequencies above the threshold frequency ($$\nu > \nu_0$$). The basic process of emission itself is independent of intensity.

Key Concepts: Linear Relationship

c) 4.5 V

[Solution Description]

To solve this problem, we use the linear relationship between stopping potential $$V_0$$ and frequency $$f$$. This can be expressed as:

$$V_0 = k \cdot f + C$$

where $$k$$ is the slope and $$C$$ is the y-intercept.

Given that at $$f = 8 \times 10^{14} \, \text{Hz}$$, $$V_0 = 3\, \text{V}$$.

Substitute these values into the equation:

$$3 = k \cdot (8 \times 10^{14}) + C \quad \text{(Equation 1)}$$

We need to find $$V_0$$ for $$f = 12 \times 10^{14} \, \text{Hz}$$:

$$V_0 = k \cdot (12 \times 10^{14}) + C$$

Since the problem states a “linear relationship” without specific constants, let’s assume the change in frequency results in proportional change in stopping potential. Assuming $$k$$ and $$C$$ remain constant, calculate the increment:

$$\Delta f = 12 \times 10^{14} – 8 \times 10^{14} = 4 \times 10^{14}$$

The increase in stopping potential due to frequency increment is proportional. Therefore,

simply applying the ratio:

$$\text{New } V_0 = 3 + (\frac{4}{8} \times 3) = 3 + 1.5 = 4.5\text{V}$$

Key Concepts: Frequency and Stopping Potential Graph, Cut-off Frequency Implication

c) Assertion is true, but Reason is false.

[Solution Description]

The assertion states that the stopping potential is greater for frequencies above the threshold frequency, which is true because the stopping potential increases with increasing frequency beyond the threshold. This is due to the relationship $$eV_0 = h\nu – \phi$$, where $$V_0$$ is the stopping potential, $$h$$ is Planck’s constant, $$\nu$$ is the frequency, and $$\phi$$ is the work function. As $$\nu$$ increases beyond the threshold frequency, the term $$h\nu$$ increases, leading to a larger $$V_0$$.

The reason given is false. The maximum kinetic energy of photoelectrons does not depend on the intensity of the incident light but rather on the frequency of the incident light, as per the equation $$K_{\text{max}} = h\nu – \phi$$. Therefore, even when the intensity changes, the maximum kinetic energy remains determined by the frequency alone.

Thus, Assertion is true, but Reason is false.

Key Concepts: Photoelectric Equation, Charge Quantization, Photon-Electron Interaction

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion discusses the relationship between the photoelectric effect and the validation of Einstein’s photoelectric equation through the quantization of charge, highlighting how the kinetic energy of emitted electrons is influenced by the frequency of incident light above a certain threshold. The reason mentions Millikan’s separate experiment on charge quantization using the oil-drop method. While both statements are true in their respective contexts, they relate to different experimental validations. The photoelectric effect addresses the quantized nature of energy absorption leading to electron emission and supports Einstein’s theoretical predictions, whereas Millikan’s experiment independently confirmed charge quantization. Hence, although both Assertion and Reason are true, the Reason does not specifically explain the Assertion as they refer to distinct phenomena and experiments.

Key Concepts: Quantum Nature, Conservation Laws

c) Assertion is true, but Reason is false.

[Solution Description]

In the photoelectric effect, photons are absorbed by electrons on a metal surface, providing the necessary energy to overcome the work function and liberate the electrons. The energy of these photons is given by $$E = h \nu$$. When an electron absorbs a photon, it gains kinetic energy equal to the photon’s energy minus the work function of the metal. The correct relation for the maximum kinetic energy $$K_{max}$$ of the ejected electrons is given by:

$$K_{max} = h\nu – \phi$$

where $$\phi$$ is the work function of the metal.

The assertion is true because energy transfer in the form of photon absorption causes electron ejection when the photon’s energy exceeds the work function. However, the reason is false because the intensity of light affects the number of photons hitting the surface per unit time, not the energy per photon or the subsequent velocity of the ejected electrons. The velocity of the electrons depends on the frequency of the incident photons, not the intensity. Thus, the intensity controls the rate of electron ejection but does not affect their velocity.

Key Concepts: Potential Difference

b) To create a potential difference
[Solution Description]
The potential difference created by the battery in a photoelectric effect experiment serves to establish an electric field that helps to direct and accelerate the emitted electrons towards the collector plate, thereby influencing the photoelectric current.

Key Concepts: Intensity and Emission

b) Intensity of light

[Solution Description] In the photoelectric effect, the number of emitted electrons is determined by the intensity of light. The light intensity affects the number of photons arriving at the surface, hence influencing the number of electrons emitted.

Key Concepts: Photon Momentum, Dual Nature

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

The assertion states that photons have momentum even though they have no rest mass. This is true because photon momentum is given by $$p = \frac{h\nu}{c}$$. The reason provided discusses the wave nature of light, which is another aspect of light’s dual nature, characterized by its particle and wave properties. Both statements are true, but the reason doesn’t explain why photons have momentum since it relates to the wave nature rather than the particle nature.

Key Concepts: Wave Theory Contradiction, Quantum Explanation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Wave theory predicts that energy is spread over the entire wavefront, meaning electrons in the metal would need time to accumulate enough energy from incoming waves to overcome the work function and be emitted. However, the photoelectric effect shows that emission occurs instantaneously once threshold frequency is exceeded, regardless of the light’s intensity. This behavior indicates that energy must be absorbed in discrete packets, or quanta, as described by quantum theory, not continuously as wave theory suggests. Therefore, while both statements are true, the Reason correctly explains why the Assertion holds because it highlights the failure of wave theory in this context.

Key Concepts: Electron Emission Processes

d) Cathode ray emission

[Solution Description]

Cathode rays are beams of electrons that are emitted when a high voltage is applied across electrodes in a vacuum tube. These rays consist of fast-moving negatively charged particles, which were later identified as electrons by J. J. Thomson. His experiments showed that these particles had specific charge $$(e/m)$$ values independent of cathode material or gas in the discharge tube.

Key Concepts: Frequency Impact

c) The maximum kinetic energy will increase.

[Solution Description] According to Einstein’s photoelectric equation $$K_{\text{max}} = h\nu – \phi_0$$, if the frequency $$\nu$$ of the incident light increases and all other factors including work function $$\phi_0$$ remain constant, the maximum kinetic energy $$K_{\text{max}}$$ of the emitted photoelectrons will increase.

Key Concepts: Advanced Application, Theoretical Explanation

c) 1.99 eV

[Solution Description]

To find the maximum kinetic energy $$(K_{\text{max}})$$ of the emitted electrons, we use Einstein’s photoelectric equation: $$K_{\text{max}} = h\nu – \phi_0$$, where $$\nu$$ is the frequency of the incident light and $$\phi_0$$ is the work function.

The work function $$\phi_0$$ can be calculated as $$h \times \nu_{\text{threshold}} = 6.626 \times 10^{-34} \times 6 \times 10^{14}$$.

$$\phi_0 = 3.9756 \times 10^{-19} \text{ J}$$

Now, calculate $$h\nu$$: $$6.626 \times 10^{-34} \times 9 \times 10^{14}$$.

$$h\nu = 5.9634 \times 10^{-19} \text{ J}$$

Finally, calculate $$K_{\text{max}}$$:

$$K_{\text{max}} = 5.9634 \times 10^{-19} – 3.9756 \times 10^{-19} = 1.9878 \times 10^{-19} \text{ J}$$