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Chapter 2: Problem 9

The wavelength associated with a golf ball weighing \(200 \mathrm{~g}\) and moving at a speed of \(5 \mathrm{~m} / \mathrm{h}\) is of the order [2001S] (a) \(10^{-10} \mathrm{~m}\) (b) \(10^{-20} \mathrm{~m}\) (c) \(10^{-30} \mathrm{~m}\) (d) \(10^{-40} \mathrm{~m}\)

Short Answer

Expert verified
The wavelength is of the order \(10^{-30}\) m, so the answer is (c).

Step by step solution

01

Convert Mass and Velocity to SI units

First, convert the mass and speed of the golf ball to SI units. The given mass is 200 grams, which needs to be converted to kilograms. Since 1 kg = 1000 g, we have: \[ \text{Mass} = \frac{200}{1000} = 0.2 \text{ kg} \]The speed is given as 5 m/h. To convert this to m/s, use the conversion factor: 1 hour = 3600 seconds. Therefore, \[ \text{Velocity} = \frac{5}{3600} \approx 0.00139 \text{ m/s} \]
02

Apply de Broglie's Wavelength Formula

De Broglie鈥檚 wavelength formula allows us to calculate the wavelength \( \lambda \) associated with any moving object:\[ \lambda = \frac{h}{mv} \]where:- \( h = 6.626 \times 10^{-34} \) Js is Planck's constant,- \( m = 0.2 \text{ kg} \) is the mass of the object,- \( v = 0.00139 \text{ m/s} \) is the velocity of the object.
03

Calculate Wavelength

Plug the values into the de Broglie wavelength formula:\[ \lambda = \frac{6.626 \times 10^{-34}}{0.2 \times 0.00139} \]Calculate the wavelength:\[ \lambda \approx \frac{6.626 \times 10^{-34}}{2.78 \times 10^{-4}} \approx 2.38 \times 10^{-30} \text{ m} \]
04

Identify the Wavelength Order

From the calculated value of \( \lambda \approx 2.38 \times 10^{-30} \text{ m} \), compare it to the answer options provided:- Option (c) is \( 10^{-30} \text{ m} \) which is closest to our calculated value,- Therefore, the wavelength is of the order \( 10^{-30} \text{ m} \).

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Key Concepts

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

Wave-Particle Duality
Wave-particle duality is a fundamental concept in quantum mechanics that describes how all particles exhibit both wave-like and particle-like properties. This dual nature was introduced to explain phenomena that classical physics could not. For instance, light behaves like a wave when it passes through a double slit, creating an interference pattern, but it can also be detected as individual particles called photons.
  • Understanding wave-particle duality helps us make sense of experiments and observations at the microscopic scale.
  • In the case of matter, such as electrons or even a golf ball, this duality implies that they have an associated wave, known as the de Broglie wavelength.
Even though macroscopic objects like golf balls have a de Broglie wavelength, it is extremely small due to their mass and speed, making wave properties negligible.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of particles at atomic and subatomic levels. It introduces concepts that significantly differ from classical mechanics, such as quantization of energy, uncertainty principles, and the aforementioned wave-particle duality.
  • These principles explain phenomena that occur on very small scales where classical laws do not apply.
  • One key idea is that particles exist in probability states until measured.
  • Quantum mechanics is essential for understanding molecular structures, semiconductor technology, and many other advanced scientific fields.
It's in this realm that the de Broglie wavelength is typically significant, illustrating how particles with mass can have characteristics of waves.
Planck's Constant
Planck's constant, denoted by the symbol \( h \), is a fundamental constant in physics that plays a critical role in the field of quantum mechanics. It is a measure of the quanta, or discrete units, of action and energy in interactions. Its value is approximately \( 6.626 \times 10^{-34} \) Joule-seconds.
  • Planck's constant is pivotal in the formula for calculating de Broglie wavelength: \( \lambda = \frac{h}{mv} \).
  • It determines the scale at which quantum effects become noticeable.
The introduction of Planck's constant marked the beginning of quantum theory and laid the groundwork for understanding the energy levels of electrons in atoms and the emission of electromagnetic radiation.

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Most popular questions from this chapter

The radius of the second Bohr orbit, in terms of the Bohr radius, \(a_{0}\), in \(\mathrm{Li}^{2+}\) is: (a) \(\frac{2 a_{0}}{3}\) (b) \(\frac{4 a_{0}}{9}\) (c) \(\frac{4 a_{0}}{3}\) (d) \(\frac{2 a_{0}}{9}\)

If \(p\) is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength \(\lambda\), then for \(1.5 p\) momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function): [Main April 8, 2019 (II)] (a) \(\frac{3}{4} \lambda\) (b) \(\frac{1}{2} \lambda\) (c) \(\frac{2}{3} \lambda\) (d) \(\frac{4}{9} \lambda\)

Consider the hydrogen atom to be a proton embedded in a cavity of radius \(a_{0}\) (Bohr radius) whose charge is neutralised by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralisation process. Also, if the magnitude of the average kinetic energy is half the magnitude of the average potential energy, find the average potential energy.

When there are two electrons in the same orbital, they have ............. spins.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is \(\left[a_{0}\right.\) is Bohr radius] : (a) \(\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}\) (b) \(\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}\) (c) \(\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}\) (d) \(\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}\)
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