Problem 54 - College Physics

Question

A specimen of the microorganism Gastropus hyptopus measures \(0.0020 \mathrm{~cm}\) in length and can swim at a speed of 2.9 times its body length per second. The tiny animal has a mass of roughly \(8.0 \times 10^{-12} \mathrm{~kg} .\) (a) Calculate the de Broglie wavelength of this organism when it is swimming at top speed. (b) Calculate the kinetic energy of the organism (in eV) when it is swimming at top speed.

Step-by-Step Solution

Answer

The de Broglie wavelength is approximately \(1.43 \times 10^{-16}\) m, and the kinetic energy is \(8.37 \times 10^{-3}\) eV.

Step 1: Calculate Velocity

The organism can swim at 2.9 times its body length per second. Given its body length is \(0.0020 \text{ cm}\), convert to meters first: \(0.0020 \text{ cm} = 0.000020 \text{ m}\). The speed \(v\) is then computed as \(v = 2.9 \times 0.000020 \text{ m/s} = 0.000058 \text{ m/s}\).

Step 2: Calculate de Broglie Wavelength

Use the de Broglie wavelength formula: \(\lambda = \frac{h}{mv}\), where \(h = 6.626 \times 10^{-34} \text{ J s}\) is Planck's constant, \(m = 8.0 \times 10^{-12} \text{ kg}\) is the organism's mass, and \(v = 0.000058 \text{ m/s}\) is the velocity. Plug in the values: \(\lambda = \frac{6.626 \times 10^{-34}}{8.0 \times 10^{-12} \times 0.000058} \approx 1.43 \times 10^{-16} \text{ meters}\).

Step 3: Calculate Kinetic Energy in Joules

The kinetic energy \(K\) is given by \(K = \frac{1}{2}mv^2\). Substitute \(m = 8.0 \times 10^{-12} \text{ kg}\) and \(v = 0.000058 \text{ m/s}\) into the equation: \(K = \frac{1}{2} \times 8.0 \times 10^{-12} \times (0.000058)^2 \approx 1.34 \times 10^{-21} \text{ J}\).

Step 4: Convert Kinetic Energy to Electron Volts

To convert joules to electron volts, use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). Hence, \(K \approx \frac{1.34 \times 10^{-21}}{1.602 \times 10^{-19}} \approx 8.37 \times 10^{-3} \text{ eV}\).

Key Concepts

Kinetic Energy Calculation

Kinetic energy is a form of energy that an object possesses due to its motion. To calculate the kinetic energy (K) of an object, you can use the formula: \[ K = \frac{1}{2}mv^2 \] where \( m \) is the mass of the object and \( v \) is the velocity. For our microorganism friend, having a mass of \( 8.0 \times 10^{-12} \text{ kg} \) and moving with a velocity of \( 0.000058 \text{ m/s} \), we plug these values into the formula:

• The mass \( (m) = 8.0 \times 10^{-12} \text{ kg} \)
• Velocity \( (v) = 0.000058 \text{ m/s} \)

This results in:\[ K = \frac{1}{2} \times 8.0 \times 10^{-12} \times (0.000058)^2 \approx 1.34 \times 10^{-21} \text{ Joules} \]Thus, the kinetic energy of this tiny organism is extremely small, highlighting how energy scales operate differently for microorganisms.

Electron Volt Conversion

Electron volts are a convenient unit of energy in the field of physics, especially when dealing with small particles like electrons. It's often more intuitive to express small quantities of energy like the kinetic energy of microorganisms in electron volts (eV) rather than in joules.To convert energy from joules to electron volts, use the conversion factor:

• \(1 ext{ eV} = 1.602 \times 10^{-19} ext{ J} \)

To express the kinetic energy of our microorganism in eV, divide its energy in joules by this conversion factor:\[ 1.34 \times 10^{-21} ext{ J} \div 1.602 \times 10^{-19} = 8.37 \times 10^{-3} ext{ eV} \]This means its kinetic energy is only a small fraction of an electron volt, reflecting the tiny scales we are dealing with when talking about microorganisms.

Planck's Constant

Planck's constant (\( h \)) is a crucial quantity in quantum mechanics that denotes the scale at which quantum effects become significant. It has a value of \( 6.626 \times 10^{-34} ext{ J s} \). This constant is key in calculations involving quantum phenomena, such as de Broglie wavelengths.When calculating the de Broglie wavelength of an object, Planck's constant helps to quantify the wave-like properties of that object:\[ \lambda = \frac{h}{mv} \]For our microorganism with mass \( 8.0 \times 10^{-12} ext{ kg} \) and velocity \( 0.000058 ext{ m/s} \), Planck’s constant allows us to determine its de Broglie wavelength to be:\[ \lambda \approx 1.43 \times 10^{-16} ext{ meters} \]This relationship illustrates the wave-particle duality of even such tiny organisms, where both their particle and wave characteristics are computed using fundamental physics principles.

Microorganism Physics

Microorganisms, due to their small size and mass, exhibit physical behaviors that bridge classical and quantum physics. When considering them in motion, features like de Broglie wavelength become applicable, even if just theoretical for most biological processes.Given the size and mass of the microorganism in our example, its de Broglie wavelength calculation imparts a quantum mechanical aspect:

• Body length: \(0.0020 ext{ cm} \) or \(0.000020 ext{ m} \)
• Traveling speed: 2.9 times its length, leading to \(0.000058 ext{ m/s} \)
• Mass: \(8.0 \times 10^{-12} ext{ kg} \)

These properties allow us to apply physics principles to calculate its de Broglie wavelength and kinetic energy, providing insights into the action of tiny organisms. This interplay enhances our understanding of how such minute entities navigate their environments, with quantum effects playing a potential role in their natural operations.