Understanding the de-Broglie Wavelength

The de-Broglie hypothesis states that all matter exhibits wave-like properties, and the wavelength associated with a particle is inversely proportional to its momentum. This wavelength is called the de-Broglie wavelength ($\lambda$). The formula for the de-Broglie wavelength is:

$$\lambda = \frac{h}{p}$$

where:

• $h$ is Planck's constant, approximately \( 6.63 \times 10^{-34} \) J s (or kg m\(^2\)/s).
• $p$ is the momentum of the particle.

For a particle with mass $m$ moving with velocity $v$, the momentum is given by $p = mv$. So, the de-Broglie wavelength is:

$$\lambda = \frac{h}{mv}$$

For a photon, which has zero rest mass, the momentum is related to energy $E$ and the speed of light $c$ by $p = E/c$. However, the de-Broglie relation $\lambda = h/p$ is universally applicable.

For a charged particle (like an electron) accelerated through a potential difference $V$, its kinetic energy (KE) is $KE = qV$, where $q$ is the charge. The kinetic energy is also related to momentum by $KE = \frac{p^2}{2m}$. Thus, $p = \sqrt{2m KE} = \sqrt{2mqV}$, and the de-Broglie wavelength is:

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

We need to calculate the de-Broglie wavelength for each case given and then arrange them in increasing order.

Calculating de-Broglie Wavelengths

Let's calculate the de-Broglie wavelength for each item (A, B, C, D, E) using the appropriate formulas. We will use Planck's constant $h = 6.63 \times 10^{-34}$ J s.

A. Bullet

Mass $m = 0.02$ kg
Speed $v = 3.3$ Km/s \( = 3.3 \times 10^3 \) m/s \( = 3300 \) m/s

Momentum $p_A = mv = (0.02 \text{ kg}) \times (3300 \text{ m/s}) = 66$ kg m/s

De-Broglie wavelength \( \lambda_A \):

$$\lambda_A = \frac{h}{p_A} = \frac{6.63 \times 10^{-34} \text{ J s}}{66 \text{ kg m/s}}$$

$$\lambda_A \approx 0.10045 \times 10^{-34} \text{ m} \approx 1.0045 \times 10^{-35} \text{ m}$$

B. Ball

Mass $m = 0.0331$ kg
Speed $v = 2$ m/s

Momentum $p_B = mv = (0.0331 \text{ kg}) \times (2 \text{ m/s}) = 0.0662$ kg m/s

De-Broglie wavelength \( \lambda_B \):

$$\lambda_B = \frac{h}{p_B} = \frac{6.63 \times 10^{-34} \text{ J s}}{0.0662 \text{ kg m/s}}$$

$$\lambda_B \approx 100.15 \times 10^{-34} \text{ m} \approx 1.0015 \times 10^{-32} \text{ m}$$

C. Dust Particle

Mass $m = 2 \times 10^{-10}$ kg
Speed $v = 3.3$ m/s

Momentum $p_C = mv = (2 \times 10^{-10} \text{ kg}) \times (3.3 \text{ m/s}) = 6.6 \times 10^{-10}$ kg m/s

De-Broglie wavelength \( \lambda_C \):

$$\lambda_C = \frac{h}{p_C} = \frac{6.63 \times 10^{-34} \text{ J s}}{6.6 \times 10^{-10} \text{ kg m/s}}$$

$$\lambda_C \approx 1.0045 \times 10^{-24} \text{ m}$$

D. Photon

Momentum $p_D = 6.63 \times 10^{-26}$ kg m/s

De-Broglie wavelength \( \lambda_D \):

$$\lambda_D = \frac{h}{p_D} = \frac{6.63 \times 10^{-34} \text{ J s}}{6.63 \times 10^{-26} \text{ kg m/s}}$$

$$\lambda_D = 1 \times 10^{-8} \text{ m}$$

E. Electron

Accelerated through a potential difference $V = 100$ V.

Mass of electron $m_e = 9.11 \times 10^{-31}$ kg (standard value)
Charge of electron $q_e = 1.602 \times 10^{-19}$ C (standard value)

Momentum $p_E = \sqrt{2m_e q_e V}$

$$p_E = \sqrt{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (1.602 \times 10^{-19} \text{ C}) \times (100 \text{ V})}$$

$$p_E = \sqrt{2 \times 9.11 \times 1.602 \times 100 \times 10^{-31} \times 10^{-19}} \text{ kg m/s}$$

$$p_E = \sqrt{292.1624 \times 10^{-50}} \text{ kg m/s}$$

$$p_E \approx 17.093 \times 10^{-25} \text{ kg m/s}$$

De-Broglie wavelength \( \lambda_E \):

$$\lambda_E = \frac{h}{p_E} = \frac{6.63 \times 10^{-34} \text{ J s}}{17.093 \times 10^{-25} \text{ kg m/s}}$$

$$\lambda_E \approx 0.3879 \times 10^{-9} \text{ m} \approx 3.879 \times 10^{-10} \text{ m}$$

Comparing the de-Broglie Wavelengths

Let's list the approximate de-Broglie wavelengths calculated:

• \( \lambda_A \approx 1.0045 \times 10^{-35} \) m (Bullet)
• \( \lambda_B \approx 1.0015 \times 10^{-32} \) m (Ball)
• \( \lambda_C \approx 1.0045 \times 10^{-24} \) m (Dust particle)
• \( \lambda_D = 1 \times 10^{-8} \) m (Photon)
• \( \lambda_E \approx 3.879 \times 10^{-10} \) m (Electron)

To arrange these in increasing order, we compare the exponents of 10 first. The smaller the exponent, the smaller the number (and the wavelength). If the exponents are the same, we compare the coefficients.

Exponents: -35, -32, -24, -10, -8.

Ordering based on exponents (from smallest to largest):

1. \( 10^{-35} \) (\(\lambda_A\))
2. \( 10^{-32} \) (\(\lambda_B\))
3. \( 10^{-24} \) (\(\lambda_C\))
4. \( 10^{-10} \) (\(\lambda_E\))
5. \( 10^{-8} \) (\(\lambda_D\))

Thus, the increasing order of de-Broglie wavelengths is:

A < B < C < E < D

This order corresponds to option 4.

Revision Table: de-Broglie Wavelength Calculations

Additional Information: de-Broglie Hypothesis and Matter Waves

The de-Broglie hypothesis, proposed by Louis de Broglie in 1924, is a fundamental concept in quantum mechanics. It suggests that just as light exhibits both wave and particle properties (wave-particle duality), matter also has wave-like characteristics. The wave associated with a material particle is called a matter wave.

Key points about the de-Broglie wavelength:

• It is inversely proportional to the momentum of the particle. Higher momentum means shorter wavelength.
• The wave nature of matter is typically observable only for microscopic particles like electrons, neutrons, and atoms, especially at low speeds, because their momentum is small, resulting in a measurable wavelength.
• For macroscopic objects like bullets or balls, the mass and velocity (and thus momentum) are large, leading to extremely small de-Broglie wavelengths (\(\sim 10^{-35}\) m or smaller), which are practically impossible to detect.
• The experimental verification of the wave nature of electrons was provided by the Davisson-Germer experiment (1927), which showed electron diffraction patterns, similar to X-ray diffraction.
• The de-Broglie concept applies to all particles, whether charged or neutral.

This question demonstrates how to calculate the de-Broglie wavelength for various objects, ranging from large, fast-moving ones to tiny quantum particles like electrons and photons, highlighting the vast range of associated wavelengths.