Problem 70 - Chemistry: An Atoms-Focused Approach

Question

How rapidly would each of the following particles be moving if they all had the same wavelength as a photon of red light \((\lambda=750 \mathrm{nm}) ?\) a. An electron of mass \(9.10938 \times 10^{-28} \mathrm{g}\) b. A proton of mass \(1.67262 \times 10^{-24} \mathrm{g}\) c. A neutron of mass \(1.67493 \times 10^{-24} \mathrm{g}\) d. An \(\alpha\) particle of mass \(6.64 \times 10^{-24} \mathrm{g}\)

Step-by-Step Solution

Answer

If each particle had the same wavelength as a photon of red light (750 nm), their speeds would be as follows: a. Electron: \(1.21 \times 10^6 \mathrm{m/s}\) b. Proton: \(5.27 \times 10^4 \mathrm{m/s}\) c. Neutron: \(5.26 \times 10^4 \mathrm{m/s}\) d. Alpha particle: \(1.32 \times 10^4 \mathrm{m/s}\)

Step 1: Identify the given information and constants

The wavelength of red light, \(\lambda\), is given as \(750 \mathrm{nm} = 750 \times 10^{-9} \mathrm{m}\). The mass of each particle is given in grams, so we need to convert the mass to kg first. Electronic mass, \(m_e = 9.10938 \times 10^{-28} \mathrm{g} = 9.10938 \times 10^{-31} \mathrm{kg}\) Proton mass, \(m_p = 1.67262 \times 10^{-24} \mathrm{g} = 1.67262 \times 10^{-27} \mathrm{kg}\) Neutron mass, \(m_n = 1.67493 \times 10^{-24} \mathrm{g} = 1.67493 \times 10^{-27} \mathrm{kg}\) Alpha particle mass, \(m_\alpha = 6.64 \times 10^{-24} \mathrm{g} = 6.64 \times 10^{-27} \mathrm{kg}\) We also need the value of Planck's constant, which is \(h = 6.626 \times 10^{-34} \mathrm{Js}\).

Step 2: Rearrange the de Broglie wavelength formula

We want to find the velocity of each particle, \(v\). Rearrange the formula to solve for \(v\): $$v = \frac{h}{m\lambda}$$

Step 3: Calculate the velocity for each particle

Now, substitute the values of \(h\), \(m\), and \(\lambda\) for each particle into the formula and calculate the velocity: a. For electron: $$v_e = \frac{6.626 \times 10^{-34}}{(9.10938 \times 10^{-31})(750 \times 10^{-9})} = 1.21 \times 10^6 \mathrm{m/s}$$ b. For proton: $$v_p = \frac{6.626 \times 10^{-34}}{(1.67262 \times 10^{-27})(750 \times 10^{-9})} = 5.27 \times 10^4 \mathrm{m/s}$$ c. For neutron: $$v_n = \frac{6.626 \times 10^{-34}}{(1.67493 \times 10^{-27})(750 \times 10^{-9})} = 5.26 \times 10^4 \mathrm{m/s}$$ d. For alpha particle: $$v_\alpha = \frac{6.626 \times 10^{-34}}{(6.64 \times 10^{-27})(750 \times 10^{-9})} = 1.32 \times 10^4 \mathrm{m/s}$$ So, if each of the particles had the same wavelength as a photon of red light, their speeds would be: a. Electron: \(1.21 \times 10^6 \mathrm{m/s}\) b. Proton: \(5.27 \times 10^4 \mathrm{m/s}\) c. Neutron: \(5.26 \times 10^4 \mathrm{m/s}\) d. Alpha particle: \(1.32 \times 10^4 \mathrm{m/s}\)

Key Concepts

Planck's constant

Planck's constant, denoted by the symbol \( h \), is a fundamental constant that plays a crucial role in quantum mechanics, specifically in the calculation of the de Broglie wavelength. It has a value of \( 6.626 \times 10^{-34} \text{J s} \), where "J" represents joules and "s" represents seconds. This constant represents the smallest action in quantum mechanics and helps bridge wave and particle-like properties of matter.

One of the primary applications of Planck's constant is in the de Broglie wavelength equation, given by: \[ \lambda = \frac{h}{mv} \] where \( \lambda \) is the wavelength, \( m \) is the mass of the particle, and \( v \) is the velocity. By using Planck's constant in this equation, it is possible to calculate the wavelength of any particle, making it pivotal in analyzing particle behavior at quantum levels.

Understanding Planck's constant is fundamental because it provides insights into the wave-particle duality of particles. At this tiny scale, matter exhibits both wave-like and particle-like properties, and this duality is crucial for comprehending phenomena such as electron diffraction and quantum tunneling.

particle velocity calculation

When calculating the velocity of particles based on their wavelength, the de Broglie wavelength formula is used to link a particle's properties to its motion. The formula can be rearranged to solve for velocity (\( v \)), given by: \[ v = \frac{h}{m\lambda} \]where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( \lambda \) is the wavelength. The challenge lies in understanding how each variable interacts and affects the calculation.

For instance, as mass \( m \) increases or wavelength \( \lambda \) decreases, the velocity \( v \) becomes smaller. Conversely, reducing the mass or increasing the wavelength results in a greater velocity. This equation shows a fascinating relationship where matter can manifest longer wavelengths despite its particulate nature.

This principle is rooted in quantum mechanics and is vital for interpreting the behaviors of particles like electrons, protons, and even elements like alpha particles. It provides valuable information on how fast these particles would move if they had the same wavelength, reflecting their potential kinetic energy derived from their wave-like attributes.

mass conversion to kg

Converting mass to kilograms is an essential step in particle physics calculations, as it ensures consistency in units and accuracy in calculations. The standard scientific unit of mass is the kilogram (kg), as per the International System of Units (SI).

When tackling problems involving small particles such as electrons, protons, and alpha particles, the masses are often given in grams (g). To convert these measurements to kilograms, you simply multiply the mass in grams by \( 10^{-3} \) because 1 gram equals 0.001 kilograms. For example:

• Electron mass: \( 9.10938 \times 10^{-28} \text{g} = 9.10938 \times 10^{-31} \text{kg} \)
• Proton mass: \( 1.67262 \times 10^{-24} \text{g} = 1.67262 \times 10^{-27} \text{kg} \)
• Alpha particle mass: \( 6.64 \times 10^{-24} \text{g} = 6.64 \times 10^{-27} \text{kg} \)

This conversion is crucial for integrating consistent units into calculations such as the de Broglie wavelength equation, where Planck's constant is typically in joules per second (J s), and wavelengths are in meters (m). Accurate mass conversion ensures the correctness of these calculations and the derived physical interpretations.