Problem 47 - Chemistry The Central Science

Question

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an 85-kg person skiing at \(50 \mathrm{km} / \mathrm{hr},\) (b) a 10.0 -g bullet fired at \(250 \mathrm{m} / \mathrm{s},\) (c) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{m} / \mathrm{s},(\mathbf{d})\) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at 550 \(\mathrm{m} / \mathrm{s}\) .

Step-by-Step Solution

Answer

The de Broglie wavelengths for the given objects are: (a) An 85-kg person skiing at 50 km/hr: \( 5.52 \times 10^{-38} \) m (b) A 10.0-g bullet fired at 250 m/s: \( 2.65 \times 10^{-34} \) m (c) A lithium atom moving at \( 2.5 \times 10^5 \) m/s: \( 2.27 \times 10^{-10} \) m (d) An ozone (\( O_3 \)) molecule moving at 550 m/s: \( 1.26 \times 10^{-15} \) m

Step 1: Convert given values to SI units

For each object, we need to convert the given mass and velocity to SI units (kilograms and meters per second respectively). (a) An 85-kg person skiing at 50 km/hr: Mass is already in kg, we need to convert the velocity from km/hr to m/s: \( v = 50 \frac{km}{hr} \times \frac{1000 m}{1 km} \times \frac{1 hr}{3600 s} = 13.89 \frac{m}{s} \) (b) A 10.0-g bullet fired at 250 m/s: We need to convert the mass from grams to kilograms: \( m = 10.0 g \times \frac{1 kg}{1000 g} = 0.01 kg \) The velocity is already in m/s. (c) A lithium atom moving at \( 2.5 \times 10^5 \) m/s: We need to find the mass of the lithium atom: 1 amu (atomic mass unit) = \( 1.66 \times 10^{-27} \) kg Lithium has an atomic mass of 7 amu. Therefore, \( m = 7 \times 1.66 \times 10^{-27} \) kg (d) An ozone (\( O_3 \)) molecule moving at 550 m/s: We need to find the mass of the ozone molecule. Oxygen has an atomic mass of 16 amu. The ozone molecule is made up of 3 oxygen atoms. Therefore, the total mass of the ozone molecule is \( 3 \times 16 = 48 \) amu. So, the mass of ozone molecule: \( m = 48 \times 1.66 \times 10^{-27} \) kg

Step 2: Calculate the de Broglie wavelengths

Now, we can use the de Broglie relationship to find the wavelengths for each object: (a) An 85-kg person skiing at 50 km/hr: \( \lambda = \frac{6.626 \times 10^{-34} \ Js}{(85 \ kg)(13.89 \frac{m}{s})} = 5.52 \times 10^{-38} \ m \) (b) A 10.0-g bullet fired at 250 m/s: \( \lambda = \frac{6.626 \times 10^{-34} \ Js}{(0.01 \ kg)(250 \frac{m}{s})} = 2.65 \times 10^{-34} \ m \) (c) A lithium atom moving at \( 2.5 \times 10^5 \) m/s: \( \lambda = \frac{6.626 \times 10^{-34} \ Js}{(7 \times 1.66 \times 10^{-27} \ kg)(2.5 \times 10^5 \frac{m}{s})} = 2.27 \times 10^{-10} \ m \) (d) An ozone (\( O_3 \)) molecule moving at 550 m/s: \( \lambda = \frac{6.626 \times 10^{-34} \ Js}{(48 \times 1.66 \times 10^{-27} \ kg)(550 \frac{m}{s})} = 1.26 \times 10^{-15} \ m \) The de Broglie wavelengths for the given objects are: (a) \( 5.52 \times 10^{-38} \) m (b) \( 2.65 \times 10^{-34} \) m (c) \( 2.27 \times 10^{-10} \) m (d) \( 1.26 \times 10^{-15} \) m

Key Concepts

Quantum Mechanics

Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It challenges classical mechanics with principles such as quantization of energy, wave-particle duality, and probabilistic nature of physical phenomena. In quantum mechanics, particles like electrons are also described as waves, a concept quantified by the de Broglie wavelength. This inherent dual character of matter allows for peculiar phenomena like tunneling, entanglement, and superposition. Quantum mechanics is not only theoretical but forms the bedrock for many modern technologies including lasers, semiconductors, and magnetic resonance imaging (MRI).

Understanding the math behind quantum mechanics, like calculating de Broglie wavelengths, is crucial because it helps us understand more about how particles behave at quantum scales. Such calculations often begin with converting values into the correct units, an essential step to ensure the accuracy of any physics-related computations.

Wave-Particle Duality

Wave-particle duality is a central concept in quantum mechanics that proposes particles of matter also have wave-like properties. This duality is best visualized through the double-slit experiment, where particles like electrons create an interference pattern, a property characteristic of waves, when not observed. However, if one tries to observe the particle, it seems to only go through one slit, displaying particle characteristics.

The de Broglie hypothesis extends this concept by suggesting that all matter has a wave-like nature. The de Broglie wavelength (\( \text{lambda} \) of a particle relates its momentum to its wave-like properties. This wavelength is given by \( \text{lambda} = \frac{h}{p} \) where \( h \) is the Planck constant and \( p \) is the momentum of the particle. The smaller the particle and the faster it moves, the shorter its de Broglie wavelength will be, making wave-like properties more prominent in the microscopic world.

SI Unit Conversion

SI units, or the International System of Units, provide a standard for measurements that facilitate consistency and accuracy in scientific communication. Conversions between different units, like those for mass and velocity, are often needed to apply formulas like the de Broglie equation.

Step-by-Step Conversions

For instance, in our exercise, we convert the velocity from km/h to meters per second (m/s) using the conversion factor \( \frac{1000 m}{1 km} \) and \( \frac{1 hr}{3600 s} \). For mass, we convert grams (g) to kilograms (kg) with the factor \( \frac{1 kg}{1000 g} \).

Importance of Correct Units

Using correct SI units is vital because the de Broglie equation requires that mass be in kilograms and velocity be in meters per second. In physics, an error in unit conversion can lead to incorrect results, which is why careful attention is crucial during this process. It is a good habit for students to always check the units in their final answer to ensure that they have correctly applied SI unit conversion principles.