Physics HOTs for Class 12 with Answers Chapter 11 Dual Nature of Radiation and Matter

1. Question:

A photon of energy $4 \, \text{eV}$ strikes a metal surface with a work function of $2 \, \text{eV}$ . If the photon is absorbed, calculate the energy of the emitted photoelectron and its maximum velocity.

Solution: The energy of the emitted photoelectron is given by:

E_{\text{kin}} = E_{\text{photon}} - \phi

Where:

$E_{\text{photon}} = 4 \, \text{eV}$
$\phi = 2 \, \text{eV}$ (work function)

Thus,

E_{\text{kin}} = 4 - 2 = 2 \, \text{eV}

To find the maximum velocity, we use the relation:

E_{\text{kin}} = \frac{1}{2}mv^2

Where $m = 9.11 \times 10^{-31} \, \text{kg}$ is the mass of the electron and $E_{\text{kin}} = 2 \, \text{eV} = 2 \times 1.6 \times 10^{-19} \, \text{J}$ .

\frac{1}{2}mv^2 = 2 \times 1.6 \times 10^{-19}

v^2 = \frac{2 \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}} = 3.51 \times 10^{11}

v \approx 5.93 \times 10^5 \, \text{m/s}

Therefore, the maximum velocity of the emitted photoelectron is approximately $5.93 \times 10^5 \, \text{m/s}$ .

2. Question:

Derive the expression for the de Broglie wavelength of a moving particle and explain its significance.

Solution: According to de Broglie, the wavelength associated with a particle is given by:

\lambda = \frac{h}{p}

Where:

$h$ is Planck’s constant ( $6.626 \times 10^{-34} \, \text{J·s}$ ),
$p = mv$ is the momentum of the particle, with $m$ being the mass and $v$ being the velocity.

Therefore, the de Broglie wavelength is:

\lambda = \frac{h}{mv}

This wavelength is significant because it connects the particle’s motion (which is traditionally described classically) with wave-like properties, showing that matter exhibits both particle and wave characteristics, fundamental to the concept of wave-particle duality.

3. Question:

What is the threshold frequency for the emission of photoelectrons from a metal surface if the work function of the metal is $2.5 \, \text{eV}$ ? Also, explain how the photoelectric effect supports the quantum nature of light.

Solution: The threshold frequency ( $f_{\text{th}}$ ) is given by the relation:

\phi = h f_{\text{th}}

Where $\phi = 2.5 \, \text{eV} = 2.5 \times 1.6 \times 10^{-19} \, \text{J}$ , and $h = 6.626 \times 10^{-34} \, \text{J·s}$ .

Solving for $f_{\text{th}}$ :

f_{\text{th}} = \frac{\phi}{h} = \frac{2.5 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 6.0 \times 10^{14} \, \text{Hz}

Explanation: The photoelectric effect demonstrates that light has a quantum nature because it shows that light can be treated as a stream of discrete energy packets called photons. The photoelectric effect supports quantum theory, as only photons with energy greater than the work function can emit electrons, indicating a particle-like behavior of light.

4. Question:

Calculate the maximum kinetic energy of the photoelectrons emitted from a surface if the frequency of the incident light is $8 \times 10^{14} \, \text{Hz}$ and the work function is $3.2 \, \text{eV}$ .

Solution: The energy of the incident photon is given by:

E_{\text{photon}} = h f

Substituting values:

E_{\text{photon}} = 6.626 \times 10^{-34} \times 8 \times 10^{14} = 5.3 \times 10^{-19} \, \text{J}

Converting this to eV:

E_{\text{photon}} = \frac{5.3 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.31 \, \text{eV}

The maximum kinetic energy of the photoelectron is given by:

E_{\text{kin}} = E_{\text{photon}} - \phi

E_{\text{kin}} = 3.31 - 3.2 = 0.11 \, \text{eV}

Therefore, the maximum kinetic energy of the emitted photoelectrons is $0.11 \, \text{eV}$ .

5. Question:

Derive the expression for the de Broglie wavelength of an electron accelerated through a potential difference $V$ .

Solution: The kinetic energy of the electron is given by:

E_k = eV

Where $e$ is the charge of the electron and $V$ is the potential difference.

The de Broglie wavelength $\lambda$ is related to the momentum $p$ by:

\lambda = \frac{h}{p}

The momentum $p$ is related to the kinetic energy by:

p = \sqrt{2mE_k} = \sqrt{2meV}

Thus, the de Broglie wavelength is:

\lambda = \frac{h}{\sqrt{2meV}}

This gives the expression for the de Broglie wavelength of an electron accelerated through a potential difference $V$ .

6. Question:

An electron is accelerated through a potential difference of $100 \, \text{V}$ . Calculate its de Broglie wavelength.

Solution: Using the formula derived earlier:

\lambda = \frac{h}{\sqrt{2meV}}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$m = 9.11 \times 10^{-31} \, \text{kg}$ ,
$e = 1.6 \times 10^{-19} \, \text{C}$ ,
$V = 100 \, \text{V}$ .

Substituting the values:

\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}}

\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2.91 \times 10^{-29}}} = 2.45 \times 10^{-11} \, \text{m}

Therefore, the de Broglie wavelength of the electron is approximately $2.45 \times 10^{-11} \, \text{m}$ .

7. Question:

Explain the concept of wave-particle duality. How does the de Broglie hypothesis provide evidence for this concept?

Solution: Wave-particle duality is the concept that every particle exhibits both wave-like and particle-like properties. The idea was proposed by Louis de Broglie, who suggested that particles such as electrons have an associated wave. The de Broglie wavelength ( $\lambda = \frac{h}{p}$ ) shows that a particle’s behavior can be described not only by its momentum but also as a wave. This is evident in phenomena such as electron diffraction, where electrons, which are traditionally considered particles, exhibit diffraction patterns characteristic of waves. This duality is a cornerstone of quantum mechanics.

8. Question:

A photon of energy $3.6 \, \text{eV}$ strikes a metal surface with a work function of $2.0 \, \text{eV}$ . Calculate the maximum kinetic energy of the emitted photoelectron.

Solution: The maximum kinetic energy is given by:

E_{\text{kin}} = E_{\text{photon}} - \phi

E_{\text{kin}} = 3.6 - 2.0 = 1.6 \, \text{eV}

Therefore, the maximum kinetic energy of the emitted photoelectron is $1.6 \, \text{eV}$ .

9. Question:

How does the photoelectric effect prove the particle nature of light?

Solution: The photoelectric effect demonstrates that light behaves as a stream of discrete packets of energy called photons. The key observation is that electrons are emitted from a metal surface only when the frequency of the incident light exceeds a certain threshold frequency, irrespective of the light’s intensity. This suggests that light energy is quantized, and only photons with energy greater than the work function of the metal can eject electrons. This result supports the particle nature of light.

10. Question:

If the work function of a metal is $1.8 \, \text{eV}$ , and a photon of energy $3.5 \, \text{eV}$ strikes the surface, calculate the stopping potential required to stop the emitted photoelectrons.

Solution: The maximum kinetic energy of the emitted photoelectrons is:

E_{\text{kin}} = E_{\text{photon}} - \phi

E_{\text{kin}} = 3.5 - 1.8 = 1.7 \, \text{eV}

The stopping potential $V_0$ is related to the maximum kinetic energy by:

E_{\text{kin}} = eV_0

V_0 = \frac{E_{\text{kin}}}{e} = 1.7 \, \text{V}

Therefore, the stopping potential required is $1.7 \, \text{V}$ .

11. Question:

A proton is accelerated through a potential difference of 200 V. Calculate its de Broglie wavelength.

Solution: The kinetic energy of the proton is given by:

E_k = eV

Where $e = 1.6 \times 10^{-19} \, \text{C}$ and $V = 200 \, \text{V}$ .

E_k = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17} \, \text{J}

The de Broglie wavelength is given by:

\lambda = \frac{h}{\sqrt{2m_p E_k}}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$m_p = 1.67 \times 10^{-27} \, \text{kg}$ (mass of proton).

Substituting values:

\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 3.2 \times 10^{-17}}}

\lambda = 5.26 \times 10^{-12} \, \text{m}

Therefore, the de Broglie wavelength of the proton is $5.26 \times 10^{-12} \, \text{m}$ .

12. Question:

How does the diffraction pattern of electrons provide evidence for their wave nature?

Solution: The diffraction pattern observed in electrons demonstrates their wave-like behavior, as diffraction is a characteristic property of waves. When electrons pass through a crystal lattice, they behave like waves, and their wavelength is given by the de Broglie relation. The interference and diffraction patterns observed are similar to those produced by light waves, providing experimental evidence for the wave nature of electrons, which supports the concept of wave-particle duality.

13. Question:

A photon of wavelength $600 \, \text{nm}$ strikes a metal surface. If the work function is $2.2 \, \text{eV}$ , calculate the maximum velocity of the emitted photoelectrons.

Solution: The energy of the incident photon is given by:

E_{\text{photon}} = \frac{hc}{\lambda}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$c = 3 \times 10^8 \, \text{m/s}$ ,
$\lambda = 600 \times 10^{-9} \, \text{m}$ .

Substituting the values:

E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = 3.31 \times 10^{-19} \, \text{J}

Converting to eV:

E_{\text{photon}} = \frac{3.31 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.07 \, \text{eV}

The maximum kinetic energy of the photoelectron is:

E_{\text{kin}} = E_{\text{photon}} - \phi = 2.07 - 2.2 = -0.13 \, \text{eV}

Since the kinetic energy is negative, no photoelectron will be emitted from this surface, as the photon energy is insufficient to overcome the work function.

14. Question:

What is the relation between the stopping potential and the kinetic energy of the emitted photoelectron?

Solution: The stopping potential $V_0$ is related to the kinetic energy of the photoelectron by the equation:

E_{\text{kin}} = eV_0

Where $E_{\text{kin}}$ is the maximum kinetic energy of the emitted electron, and $e$ is the charge of the electron. Therefore, the stopping potential is the potential required to stop the most energetic photoelectron emitted during the photoelectric effect.

15. Question:

An electron has a de Broglie wavelength of $2 \, \text{nm}$ . Calculate its momentum.

Solution: Using de Broglie’s relation:

\lambda = \frac{h}{p}

Where:

$\lambda = 2 \times 10^{-9} \, \text{m}$ ,
$h = 6.626 \times 10^{-34} \, \text{J·s}$ .

Solving for $p$ :

p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{2 \times 10^{-9}} = 3.31 \times 10^{-25} \, \text{kg·m/s}

16. Question:

What is the energy of a photon emitted when an electron transitions from $n = 4$ to $n = 2$ in a hydrogen atom?

Solution: The energy difference between two levels is given by:

E = -13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \, \text{eV}

For $n_1 = 2$ and $n_2 = 4$ :

E = -13.6 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = -13.6 \left(\frac{1}{4} - \frac{1}{16}\right) = -13.6 \times \frac{3}{16} = 2.55 \, \text{eV}

Therefore, the energy of the emitted photon is $2.55 \, \text{eV}$ .

17. Question:

How does the concept of wave-particle duality explain the behavior of electrons in an atom?

Solution: Wave-particle duality explains that electrons, traditionally considered as particles, also exhibit wave-like properties. This duality is crucial in understanding electron behavior in atoms. Electrons are confined to discrete energy levels in atoms because their wave-like nature allows only certain wavelengths to fit within the orbit, leading to the quantization of energy levels. The electron behaves both as a particle (for momentum and position) and as a wave (for its energy and the interaction with other particles), which is fundamental to quantum mechanics.

18. Question:

A photon of energy $2.5 \, \text{eV}$ strikes a metal surface. If the work function is $1.8 \, \text{eV}$ , calculate the stopping potential required to stop the emitted photoelectrons.

Solution: The maximum kinetic energy of the photoelectrons is:

E_{\text{kin}} = E_{\text{photon}} - \phi = 2.5 - 1.8 = 0.7 \, \text{eV}

The stopping potential $V_0$ is related to the kinetic energy by:

E_{\text{kin}} = eV_0

V_0 = \frac{0.7}{1} = 0.7 \, \text{V}

19. Question:

How does the photoelectric effect provide evidence for the quantum nature of light?

Solution: The photoelectric effect demonstrates that light behaves as a stream of discrete energy packets called photons. The key observations are:

Photoelectrons are emitted only when the frequency of light exceeds a certain threshold, irrespective of its intensity, indicating that the energy of light is quantized.
The kinetic energy of the emitted photoelectrons depends on the frequency of the incident light, not its intensity, confirming that the energy of each photon is proportional to its frequency.

This supports the quantum nature of light, as described by Einstein.

20. Question:

Calculate the wavelength of an electron moving with a velocity of $2 \times 10^6 \, \text{m/s}$ .

Solution: Using de Broglie's formula:

\lambda = \frac{h}{mv}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$m = 9.11 \times 10^{-31} \, \text{kg}$ (mass of the electron),
$v = 2 \times 10^6 \, \text{m/s}$ .

Substituting values:

\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2 \times 10^6}

\lambda = 3.64 \times 10^{-10} \, \text{m}

21. Question:

Prove that the de Broglie wavelength of a particle is inversely proportional to its momentum.

Solution: de Broglie proposed that every moving particle has an associated wavelength, known as the de Broglie wavelength. The relation is given by:

\lambda = \frac{h}{p}

Where:

$\lambda$ is the de Broglie wavelength,
$h$ is Planck's constant ( $6.626 \times 10^{-34} \, \text{J·s}$ ),
$p$ is the momentum of the particle.

Since momentum $p$ is the product of the mass $m$ and velocity $v$ of the particle:

p = mv

Therefore, the de Broglie wavelength becomes:

\lambda = \frac{h}{mv}

Hence, we can conclude that the de Broglie wavelength $\lambda$ is inversely proportional to the momentum $p$ .

22. Question:

An electron is moving with a velocity of $1 \times 10^6 \, \text{m/s}$ . Calculate its de Broglie wavelength and compare it with the wavelength of visible light.

Solution: First, we calculate the de Broglie wavelength using the formula:

\lambda = \frac{h}{mv}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$m = 9.11 \times 10^{-31} \, \text{kg}$ (mass of the electron),
$v = 1 \times 10^6 \, \text{m/s}$ .

Substituting values:

\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 1 \times 10^6}

\lambda = 7.28 \times 10^{-11} \, \text{m}

The wavelength of visible light is in the range of $400 \, \text{nm}$ to $700 \, \text{nm}$ ( $4 \times 10^{-7} \, \text{m}$ to $7 \times 10^{-7} \, \text{m}$ ).

Clearly, the de Broglie wavelength of the electron ( $7.28 \times 10^{-11} \, \text{m}$ ) is much smaller than that of visible light, highlighting the wave-particle duality of matter at microscopic scales.

23. Question:

How does the de Broglie wavelength of a macroscopic object compare to that of an electron?

Solution: The de Broglie wavelength of a macroscopic object is given by:

\lambda = \frac{h}{mv}

For large macroscopic objects, the mass $m$ and velocity $v$ are much larger than those of microscopic particles like electrons. As a result, the de Broglie wavelength of macroscopic objects is extremely small.

For example, for a car of mass $1000 \, \text{kg}$ moving at $10 \, \text{m/s}$ :

\lambda = \frac{6.626 \times 10^{-34}}{1000 \times 10} = 6.626 \times 10^{-38} \, \text{m}

This is far smaller than atomic dimensions, making wave-like behavior for macroscopic objects undetectable. This explains why wave properties are only observable for particles at microscopic scales.

24. Question:

An electron with a de Broglie wavelength of $1.5 \, \text{nm}$ is accelerated by a potential difference of $1000 \, \text{V}$ . Find the velocity of the electron.

Solution: The de Broglie wavelength of the electron is given by:

\lambda = \frac{h}{mv}

Rearranging to find the velocity:

v = \frac{h}{m \lambda}

Substituting values:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$m = 9.11 \times 10^{-31} \, \text{kg}$ ,
$\lambda = 1.5 \times 10^{-9} \, \text{m}$ .

v = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.5 \times 10^{-9}} = 4.87 \times 10^6 \, \text{m/s}

25. Question:

Calculate the energy of a photon of wavelength $400 \, \text{nm}$ .

Solution: The energy of a photon is given by:

E = \frac{hc}{\lambda}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$c = 3 \times 10^8 \, \text{m/s}$ ,
$\lambda = 400 \times 10^{-9} \, \text{m}$ .

Substituting values:

E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J}

Converting to eV:

E = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.11 \, \text{eV}

26. Question:

If the work function of a metal is $2.5 \, \text{eV}$ and light of wavelength $500 \, \text{nm}$ falls on it, calculate the maximum kinetic energy of the emitted photoelectron.

Solution: The energy of the incident photon is:

E_{\text{photon}} = \frac{hc}{\lambda}

Where:

$h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
$c = 3 \times 10^8 \, \text{m/s}$ ,
$\lambda = 500 \times 10^{-9} \, \text{m}$ .

Substituting values:

E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \, \text{J}

Converting to eV:

E_{\text{photon}} = \frac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.49 \, \text{eV}

The maximum kinetic energy of the emitted photoelectron is:

E_{\text{kin}} = E_{\text{photon}} - \phi = 2.49 - 2.5 = -0.01 \, \text{eV}

Since the kinetic energy is negative, no photoelectron will be emitted because the energy of the photon is insufficient to overcome the work function.

27. Question:

Explain why the photoelectric effect cannot be explained by the wave theory of light.

Solution: The wave theory of light predicts that the energy of light should depend on its intensity, not its frequency. According to the wave theory, the energy delivered by light to the electrons should accumulate over time, and photoelectrons should be emitted after a delay, if the light's intensity is high enough, even if the frequency is below the threshold. However, experiments show that photoelectrons are emitted instantaneously when light of a frequency above the threshold is shone on the metal, and no electrons are emitted below this frequency, regardless of the light's intensity. This cannot be explained by the wave theory and led to the quantum theory of light, where light behaves as discrete packets of energy (photons).

28. Question:

What is the significance of the de Broglie wavelength for microscopic particles like electrons and atoms?

Solution: The de Broglie wavelength provides a means of understanding the wave-like nature of particles, particularly at microscopic scales. For particles like electrons and atoms, the wavelength is significant because it allows these particles to exhibit interference and diffraction effects, phenomena typically associated with waves. This wave-particle duality explains various quantum phenomena such as the behavior of electrons in atoms and the electron diffraction patterns observed in experiments. For macroscopic objects, however, the de Broglie wavelength is exceedingly small, making wave-like behavior unobservable.

29. Question:

Discuss the concept of wave-particle duality using the example of electron diffraction.

Solution: Wave-particle duality is the concept that every particle exhibits both wave-like and particle-like properties. In the case of electron diffraction, when electrons are accelerated and directed at a crystal, they exhibit diffraction patterns typically associated with waves. This was demonstrated in the Davisson-Germer experiment, where electrons behaved like waves when passing through a crystal, leading to interference patterns. This confirmed the wave nature of matter, as predicted by de Broglie. The observation of diffraction patterns in electron beams is evidence of their wave-like behavior, showing that particles like electrons can also exhibit wave-like characteristics.

30. Question:

Calculate the threshold frequency of a metal if its work function is $4.2 \, \text{eV}$ .

Solution: The threshold frequency $f_0$ is related to the work function $\phi$ by the equation:

\phi = h f_0

Rearranging to find $f_0$ :

f_0 = \frac{\phi}{h}

Converting the work function to joules:

\phi = 4.2 \, \text{eV} = 4.2 \times 1.6 \times 10^{-19} \, \text{J} = 6.72 \times 10^{-19} \, \text{J}

Now, calculating $f_0$ :

f_0 = \frac{6.72 \times 10^{-19}}{6.626 \times 10^{-34}} = 1.01 \times 10^{15} \, \text{Hz}

Dual Nature of Radiation and Matter

Class 12th Physics Chapter HOTs

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Class 12^th Physics Chapter HOTs