Calculating De-Broglie Wavelength of a Particle

The question asks for the de-Broglie wavelength of a particle with a given mass and velocity. The de-Broglie hypothesis states that all matter exhibits wave-like properties, and the wavelength associated with a moving particle is called the de-Broglie wavelength.

The de-Broglie wavelength ($\lambda$) is related to the momentum ($p$) of the particle by the following formula:

\(\lambda = \frac{h}{p}\)

Where:

• \(\lambda\) is the de-Broglie wavelength
• \(h\) is Planck's constant
• \(p\) is the momentum of the particle

Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$).

\(p = mv\)

Substituting the momentum formula into the de-Broglie wavelength formula, we get:

\(\lambda = \frac{h}{mv}\)

Now, let's use the given values from the question:

• Mass of the particle, \(m = 0.001 \text{ kg}\)
• Velocity of the particle, \(v = 100 \text{ m/s}\)
• Planck's constant, \(h \approx 6.626 \times 10^{-34} \text{ J s}\) (Using \(6.62 \times 10^{-34} \text{ J s}\) as indicated by the options)

Let's plug these values into the formula:

\(\lambda = \frac{6.62 \times 10^{-34} \text{ J s}}{(0.001 \text{ kg}) \times (100 \text{ m/s})}\)

First, calculate the momentum ($mv$):

\(mv = 0.001 \text{ kg} \times 100 \text{ m/s} = 0.1 \text{ kg m/s}\)

Now, calculate the de-Broglie wavelength:

\(\lambda = \frac{6.62 \times 10^{-34} \text{ J s}}{0.1 \text{ kg m/s}}\)

\(\lambda = \frac{6.62 \times 10^{-34}}{10^{-1}} \text{ m}\)

\(\lambda = 6.62 \times 10^{-34} \times 10^{1} \text{ m}\)

\(\lambda = 6.62 \times 10^{(-34 + 1)} \text{ m}\)

\(\lambda = 6.62 \times 10^{-33} \text{ m}\)

Comparing this result with the given options, we find that the calculated de-Broglie wavelength is \(6.62 \times 10^{-33} \text{ m}\).

Therefore, the correct option is \(6.62 \times 10^{-33} \text{ m}\).

Revision Table: Key Concepts

Additional Information on De-Broglie Wavelength

The de-Broglie hypothesis, proposed by Louis de Broglie in 1924, revolutionized our understanding of matter. It suggested that particles, like electrons, protons, or even everyday objects, can exhibit wave-like behavior, just as waves (like light) can exhibit particle-like behavior (photons).

• Wave-Particle Duality: The concept that particles can behave like waves and waves can behave like particles is known as wave-particle duality. It's a cornerstone of quantum mechanics.
• Significance: The de-Broglie wavelength is significant primarily for microscopic particles like electrons, where the wavelength is comparable to the size of atoms or interatomic distances. This is why electron diffraction experiments provide strong evidence for the wave nature of electrons.
• For macroscopic objects: For everyday objects like the 0.001 kg particle in this problem, even though it has a de-Broglie wavelength, the value is extremely small (\(10^{-33} \text{ m}\)). This wavelength is far too small to be observed or detected by any current experimental method, which is why we don't observe wave-like behavior for macroscopic objects in our daily lives.
• Relation to Kinetic Energy: The de-Broglie wavelength can also be expressed in terms of the kinetic energy (\(KE\)) of the particle. Since \(KE = \frac{1}{2}mv^2\) and \(p = mv\), we have \(p^2 = m^2v^2 = 2m (\frac{1}{2}mv^2) = 2m KE\). Thus, \(p = \sqrt{2m KE}\). The de-Broglie wavelength becomes \(\lambda = \frac{h}{\sqrt{2m KE}}\).