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Photon of frequency \(v\) has a momentum associated with it. If \(c\) is the velocity of the light, the moments is (a) \(\frac{v}{c}\) (b) \(h v c\) (c) \(\frac{h v}{c^{2}}\) (d) \(\frac{h v}{c}\)

Short Answer

Expert verified
(d) \( \frac{h v}{c} \)

Step by step solution

01

Understand the Question

The problem asks us to determine the expression for the momentum of a photon with a frequency \( v \). We know that light behaves as both a wave and a particle, so we need to consider its particle nature here.
02

Use the Momentum Formula for Photons

The momentum \( p \) of a photon is given by the formula: \[ p = \frac{h v}{c} \] where \( h \) is Planck's constant, \( v \) is the frequency of the photon, and \( c \) is the speed of light.
03

Match with the Given Options

Looking at the formula \( \frac{h v}{c} \), we compare this to the provided options: (a) \( \frac{v}{c} \) (b) \( h v c \) (c) \( \frac{h v}{c^{2}} \) (d) \( \frac{h v}{c} \)The formula \( \frac{h v}{c} \) matches option (d).
04

Verify the Reasoning

Confirm that the formula \( \frac{h v}{c} \) correctly represents the momentum of a photon and matches one of the options. Photons do have momentum despite having no rest mass, and this expression correctly follows from the dual wave-particle nature of light.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planck's Constant
Planck's Constant is a fundamental element in quantum mechanics. It bridges the world of waves and particles. This constant is denoted by the letter \( h \), and has a value of approximately \( 6.63 \times 10^{-34} \text{Js} \). It was first introduced by Max Planck, who was investigating black-body radiation.
Understanding Planck's Constant is crucial because it quantifies the smallest action possible in quantum mechanics. It's involved in equations that describe the energy of photons, where energy \( E \) is equals to \( E = h u \), linking frequency \( u \) to energy.

The constant plays a central role in the formula for photon momentum. Here, it is part of the expression \( \frac{h u}{c} \), indicating that photon momentum is directly proportional to frequency \( u \). Thus, without Planck's constant, we wouldn't be able to accurately calculate the momentum of a photon or understand atomic scale phenomena.
Wave-Particle Duality
Wave-Particle Duality is a core concept in modern physics that explains the dual behavior of light and matter. It asserts that particles such as photons exhibit both wave and particle characteristics. This duality was experimentally established through phenomena such as the double-slit experiment and the photoelectric effect.

When we speak about photons in terms of momentum, we're considering their particle characteristics. But photons naturally also display wave-like properties, such as interference and diffraction. This dual nature is why the formula for photon momentum couples concepts from both classical wave theory and particle physics.

In this particular exercise, the use of frequency in calculating photon momentum echoes this duality. Frequency is a property typically associated with waves, yet it's integral to understanding photon behavior as a particle. This is what makes the dual nature of photons a pivotal topic for any physics enthusiast.
Speed of Light
The Speed of Light, denoted as \( c \), is a crucial constant in physics. It represents the maximum speed at which all energy, matter, and information in the universe can travel. In a vacuum, the speed of light is approximately \( 3 \times 10^8 \text{ m/s} \). This speed is not just a number but a cornerstone of Einstein's Theory of Relativity, impacting how we understand time and space.

In the context of photon momentum, the speed of light appears in the denominator of the expression \( \frac{h u}{c} \). This reflects the profound relationship between energy, light speed, and momentum. Even though photons have no mass, their momentum depends on how they travel through space at light speed. It's not just their energy \( hv \) that matters; their velocity \( c \) also plays a critical role.

By understanding the speed of light, students gain insight into why photons, despite being massless, are such effective carriers of energy and momentum. Conceptualizing the speed of light helps unravel the deeper insights of our universe, aided by the equations it inhabits.

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Most popular questions from this chapter

A proton, a neutron, an electron and an \(\alpha\)-particle have same energy. Then their de-Broglie wavelengths compare as [NCERT Exemplar] (a) \(\lambda_{\mathrm{p}}=\lambda_{\mathrm{n}}>\lambda_{\mathrm{e}}>\lambda_{\alpha}\) (b) \(\lambda_{\alpha}=\lambda_{\mathrm{p}}>\lambda_{\mathrm{n}}>\lambda_{e}\) (c) \(\lambda_{e}=\lambda_{\mathrm{p}}>\lambda_{\mathrm{n}}>\lambda_{\alpha}\) (d) \(\lambda_{e}=\lambda_{p}>\lambda_{n}>\lambda_{a}\)

A metal surface of work function \(1.07 \mathrm{eV}\) is irradiated with light of wavelength \(332 \mathrm{~nm}\). The retarding potential required to stop the escape of photoelectrons is (a) \(1.07 \mathrm{eV}\) (b) \(2.66 \mathrm{eV}\) (c) \(3.7 \mathrm{eV}\) (d) \(4.81 \mathrm{eV}\)

Two monochromatic beams \(A\) and \(B\) of equal intensity \(I\), hit a screen. The number of photons hitting the screen by beam \(A\) is twice that by beam \(B\). Then, what inference can you move about their frequencies? (a) The frequency of beam \(B\) is twice that of \(A\) (b) The frequency of beam \(B\) is half that of \(A\) (c) The frequency of beam \(A\) is twice of \(B\) (d) None of the above

A \(100 \mathrm{~W}\) light bulb is placed at the centre of a spherical chamber of radius \(0.10 \mathrm{~m} .\) Assume that \(66 \%\) of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is (a) \(0.87 \times 10^{-6} \mathrm{~Pa}\) (b) \(1.77 \times 10^{-6} \mathrm{~Pa}\) (c) \(3.50 \times 10^{-6} \mathrm{~Pa}\) (d) None of the above

A \(100 \mathrm{~W}\) 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 \(589 \mathrm{~nm} .\) (i) What is the energy per photon associated with the sodium light? (ii) At what rate are the photons delivered to the sphere? (a) (i) \(4.6 \mathrm{eV}\) (ii) \(1.6 \times 10^{24}\) photon/s (b) (i) \(3.4 \mathrm{eV}\) (ii) \(4.5 \times 10^{24}\) photon/s (c) (i) \(2.1 \mathrm{eV}\) (ii) \(3 \times 10^{20}\) photon/s (d) (i) \(1.1 \mathrm{eV}\) (ii) \(2 \times 10^{24}\) photon/s

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