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Chapter 25: Problem 4

A particle of mass \(1 \mathrm{mg}\) has the same wavelength as an electron moving with a velocity of \(3 \times 10^{6} \mathrm{~ms}^{-1}\). The velocity of the particle is (a) \(3 \times 10^{-31} \mathrm{~ms}^{-1}\) (b) \(2.7 \times 10^{-21} \mathrm{~ms}^{-1}\) (c) \(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\) (d) \(9 \times 10^{-2} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity of the particle is approximately \(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\), which is option (c).

Step by step solution

01

Understanding De Broglie's Equation

De Broglie's hypothesis states that every moving particle or object has an associated wave. The wavelength \( \lambda \) is given by the equation \( \lambda = \frac{h}{p} \) where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \mathrm{~m}^2\mathrm{~kg}~\mathrm{s}^{-1} \) and \( p \) is the momentum (product of mass and velocity) of the particle.
02

Calculate Wavelength for the Electron

To find the wavelength of the electron, use the formula \( \lambda = \frac{h}{mv} \). For the electron, use its mass \( m_e \approx 9.11 \times 10^{-31} \mathrm{~kg} \) and its velocity \( v_e = 3 \times 10^{6} \mathrm{~ms}^{-1} \). Substituting these into the formula gives: \[ \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^6} \approx 2.43 \times 10^{-10} \mathrm{~m} \]
03

Equate Wavelengths for Particle and Electron

Since the particle and the electron have the same wavelength, the equation for both is equal. Therefore, setting the wavelengths equal gives us:\[ \frac{h}{m_p v_p} = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^6} \] where \( m_p = 1 \times 10^{-3} \mathrm{~g} = 1 \times 10^{-6} \mathrm{~kg} \).
04

Solve for Velocity of the Particle

Rearange the equation to solve for the velocity \( v_p \) of the particle:\[ v_p = \frac{h}{m_p \times 2.43 \times 10^{-10}} \] Substitute \( h = 6.626 \times 10^{-34} \mathrm{~m}^2\mathrm{~kg}~\mathrm{s}^{-1} \) and \( m_p = 1 \times 10^{-6} \mathrm{~kg} \) to find \( v_p \):\[ v_p = \frac{6.626 \times 10^{-34}}{1 \times 10^{-6} \times 2.43 \times 10^{-10}} \approx 2.73 \times 10^{-18} \mathrm{~ms}^{-1} \]
05

Choose the Correct Answer from Options

The calculated velocity \( v_p \approx 2.73 \times 10^{-18} \mathrm{~ms}^{-1} \) matches option (c) \(2.7 \times 10^{-18} \mathrm{~ms}^{-1}\). Therefore, the correct answer is (c).

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Key Concepts

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

Wavelength of particles
In the world of quantum mechanics, the concept of wavelength is not just limited to waves like sound or light. According to De Broglie's hypothesis, particles can also behave like waves, and thus, have an associated wavelength. This fundamental idea proposes that every moving object or particle has a wave nature.
To understand this better, consider a particle with mass and velocity. Its associated wavelength, known as the De Broglie wavelength, is inversely proportional to its momentum. The larger the momentum, the shorter the wavelength. This relationship is central to explaining the behavior of particles on a quantum level.
For instance, electrons, which are typically associated with wave properties due to their small mass and high speed, exhibit significant wavelengths that are measurable and impactful in experiments.
Momentum and mass
Momentum is a crucial concept in physics, representing the quantity of motion an object has. It is calculated as the product of an object's mass and velocity. This means that as either mass or velocity increases, so does the momentum.
In De Broglie's equation, momentum is pivotal. The equation \( \lambda = \frac{h}{p} \) shows that the wavelength \( \lambda \) is inversely related to the momentum \( p \), which is derived from multiplying mass and velocity. As a particle's momentum increases, its De Broglie wavelength decreases, illustrating an elegant link between classical and quantum physics.
For the problem above, understanding how mass affects momentum helps explain why heavier particles moving at similar speeds as lighter ones tend to have smaller wavelengths.
Planck's constant
Planck's constant, denoted as \( h \), is a fundamental quantity in quantum mechanics, acting as a bridge between the particle and wave nature of matter. Its value is approximately \( 6.626 \times 10^{-34} \mathrm{~m}^2\mathrm{~kg}~\mathrm{s}^{-1} \), and it is crucial in the calculation of the De Broglie wavelength.
The constant appears in multiple key equations in quantum mechanics, including the well-known equation \( E = hf \), which relates energy \( E \) to the frequency \( f \) of a photon.
In the context of De Broglie's hypothesis, Planck's constant normalizes the momentum to give a meaningful wavelength. This allows scientists to predict and understand the duality of particles more effectively, leading to breakthroughs in understanding atomic and subatomic systems.

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Most popular questions from this chapter

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

Radiations of two photon's energy, twice and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of photoelectrons emitted in two cases is (a) \(1: 2\) (b) \(1: 3\) (c) \(1: 4\) (d) \(1: 1\)

In a photoelectric experiment for 4000 \AA incident radiation. The potential difference to stop the ejection is \(2 \mathrm{~V}\). If the incident light is changed to \(3000 \AA\), then the potential required to stop the ejection of electrons will be (a) greater than \(2 \mathrm{~V}\) (b) less than \(2 \mathrm{~V}\) (c) \(\infty\) (d) zero

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

There are two sources of light each emitting with a power of \(100 \mathrm{~W}\). One emits X-rays of wavelength \(1 \mathrm{~nm}\) and the other visible light of wavelength \(500 \mathrm{~nm} .\) Find the ratio of number of photons of X-rays in the photons of visible light of the given wavelength? (a) \(1: 500\) (b) \(1: 250\) (c) \(1: 20\) (d) 100
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