Problem 26 - Chemistry and Chemical Reactivity

Question

A rifle bullet (mass \(=1.50 \mathrm{g}\) ) has a velocity of \(7.00 \times 10^{2} \mathrm{mph}\) (miles per hour). What is the wavelength associated with this bullet?

Step-by-Step Solution

Answer

The wavelength associated with the bullet is approximately \(1.41 \times 10^{-33}\) meters.

Step 1: Convert Units

First, we convert the bullet's velocity from miles per hour to meters per second. 1 mile = 1609.34 meters, 1 hour = 3600 seconds, Thus the conversion is \[ 700 \text{ mph} \approx 700 \times \frac{1609.34}{3600} \approx 312.85 \text{ m/s} \]

Step 2: Convert Mass Units

Next, convert the mass from grams to kilograms since SI units require kilograms.1 gram = 0.001 kilograms. Thus,\[ 1.50 \text{ g} = 1.50 \times 0.001 = 0.0015 \text{ kg} \]

Step 3: Use de Broglie Wavelength Formula

Using the de Broglie equation, which states that the wavelength \(\lambda\) is given by \[ \lambda = \frac{h}{mv} \]where \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ J s}\), \(m\) is mass (in kg), and \(v\) is velocity (in m/s).

Step 4: Calculate the Wavelength

Substitute the given mass and velocity into the de Broglie formula:\[ \lambda = \frac{6.63 \times 10^{-34}}{0.0015 \times 312.85} \]Calculate \(mv = 0.0015 \times 312.85 = 0.469275 \text{ kg m/s} \) Thus,\[ \lambda \approx \frac{6.63 \times 10^{-34}}{0.469275} \approx 1.41 \times 10^{-33} \text{ meters} \]

Step 5: Interpret the Result

The calculated wavelength \(1.41 \times 10^{-33}\) meters is extremely small, which is consistent with the fact that macroscopic objects like bullets have very short wavelengths that do not exhibit noticeable wave properties.

Key Concepts

Velocity Conversion

When dealing with problems involving different units, converting units is often the first step. In this exercise, we need to convert the bullet's velocity from miles per hour (mph) to meters per second (m/s). This process involves knowing that one mile is equivalent to 1609.34 meters and one hour has 3600 seconds. By multiplying the original velocity, 700 mph, by the conversion factors \(\frac{1609.34}{3600}\), we can calculate the speed in SI units.
This results in approximately 312.85 m/s. Velocity conversion is crucial because it enables us to use consistent units when applying formulas, such as the de Broglie equation, which demands SI units for speed.

Mass Conversion

Converting the mass from grams to kilograms is essential for working with the de Broglie wavelength. The de Broglie equation requires SI units, where mass is expressed in kilograms. The standard conversion is 1 gram equals 0.001 kilograms. Thus, you convert 1.50 grams to kilograms by multiplying by 0.001.
This simple step changes our mass to 0.0015 kilograms, ensuring all units in our equations are consistent with SI standards. Accurately converting mass is key to achieving correct calculations in physics problems.

Planck's Constant

Planck's constant is a fundamental constant used in quantum mechanics, symbolized as \(h\). It links the energy of photons to the frequency of their electromagnetic waves. In this problem, its value is \(6.63 \times 10^{-34}\) joule seconds, which is a very small number.

• Planck's constant is a cornerstone of quantum mechanics, reflecting the quantized nature of the universe.
• It plays a vital role in the de Broglie wavelength formula \(\lambda = \frac{h}{mv}\), where \(m\) is mass and \(v\) is velocity.

This constant helps us find the extremely short wavelengths of macroscopic objects like bullets.

SI Units

SI units or the International System of Units is a globally accepted system for measuring various physical quantities. The importance of using SI units in scientific equations cannot be understated.

• They provide a standard and consistency needed for clear, accurate calculations.
• SI units for mass are kilograms, length in meters, and time in seconds.

In this exercise, we saw both mass and velocity converted into SI units before applying them to the de Broglie formula. Consistent use of SI units ensures that we can properly and accurately apply universal physics equations to numerous scenarios.