Question

Compare the wavelengths of an electron (mass \(=9.11 \times 10^{-31} \mathrm{~kg}\) ) and a proton (mass \(=1.67 \times 10^{-27} \mathrm{~kg}\) ), each having (a) a speed of \(3.4 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a kinetic energy of \(2.7 \times 10^{-15} \mathrm{~J}\)

Short answer

Electrons have higher wavelength than protons at the same speed and kinetic energy.

Step-by-step solution

Understand the relationship between wavelength and momentum

The wavelength \( \lambda \) of a particle can be found using De Broglie's equation: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \ \mathrm{Js} \)), and \( p \) is the momentum of the particle.

Calculate momentum for each particle with given speed

Momentum \( p \) is given by \[ p = mv \] where \( m \) is the mass and \( v \) is the velocity. For the electron, \[ p_{e} = (9.11 \times 10^{-31} \ \mathrm{kg})(3.4 \times 10^{6} \ \mathrm{m/s}) = 3.10 \times 10^{-24} \ \mathrm{kg \cdot m/s} \] For the proton, \[ p_{p} = (1.67 \times 10^{-27} \ \mathrm{kg})(3.4 \times 10^{6} \ \mathrm{m/s}) = 5.68 \times 10^{-21} \ \mathrm{kg \cdot m/s} \]

Compute the wavelength using De Broglie's equation

Using the momenta calculated in Step 2, we can now find the wavelengths. For the electron, \[ \lambda_{e} = \frac{h}{p_{e}} = \frac{6.626 \times 10^{-34}}{3.10 \times 10^{-24}} = 2.14 \times 10^{-10} \ \mathrm{m} \] For the proton, \[ \lambda_{p} = \frac{h}{p_{p}} = \frac{6.626 \times 10^{-34}}{5.68 \times 10^{-21}} = 1.17 \times 10^{-13} \ \mathrm{m} \]

Calculate wavelength with given kinetic energy

The kinetic equation \( K = \frac{1}{2}mv^2 \) rearranges to \[ v = \sqrt{ \frac{2K}{m}} \] Next, calculate velocity for each particle. For the electron, \[ v_{e} = \sqrt{ \frac{2 \times 2.7 \times 10^{-15}}{9.11 \times 10^{-31}}} = 2.45 \times 10^7 \ \mathrm{m/s} \] For the proton, \[ v_{p} = \sqrt{ \frac{2 \times 2.7 \times 10^{-15}}{1.67 \times 10^{-27}}} = 1.80 \times 10^6 \ \mathrm{m/s} \]

Compute the wavelength for each particle with the velocities from kinetic energy

Again, find their momenta. For the electron, \[ p_{e} = (9.11 \times 10^{-31} \ \mathrm{kg})(2.45 \times 10^{7} \ \mathrm{m/s}) = 2.23 \times 10^{-23} \ \mathrm{kg \cdot m/s} \] For the proton, \[ p_{p} = (1.67 \times 10^{-27} \ \mathrm{kg})(1.80 \times 10^{6} \ \mathrm{m/s}) = 3.01 \times 10^{-21} \ \mathrm{kg \cdot m/s} \] Finally, use De Broglie's equation, For the electron, \[ \lambda_{e} = \frac{6.626 \times 10^{-34}}{2.23 \times 10^{-23}} = 2.97 \times 10^{-11} \ \mathrm{m} \] For the proton, \[ \lambda_{p} = \frac{6.626 \times 10^{-34}}{3.01 \times 10^{-21}} = 2.20 \times 10^{-13} \ \mathrm{m} \]

Key concepts

De Broglie's equation

De Broglie's equation forms the cornerstone of understanding how particles behave like waves, which is a key idea in quantum mechanics. This equation links a particle's wavelength, known as the 'de Broglie wavelength', to its momentum, and is given by \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \ \mathrm{Js} \)), and \( p \) is the particle's momentum.
This equation shows that even particles with mass, like electrons and protons, have wave properties. The smaller the momentum, the larger the wavelength, and vice versa. This inverse relationship explains why particles of different masses traveling at the same speed will have different wavelengths, as we will see in the exercise.
Understanding de Broglie's equation is crucial for grasping the dual nature of particles and waves in quantum physics.

Momentum

Momentum is a fundamental concept in both classical and quantum physics. It is defined as the product of an object's mass (m) and velocity (v), given by the formula \[ p = mv \]
In the context of the exercise, we calculate the momentum for an electron and a proton given their masses and velocities. For example, the momentum of an electron with a mass of \( 9.11 \times 10^{-31} \ \mathrm{kg} \) and a velocity of \( 3.4 \times 10^{6} \ \mathrm{m/s} \) is calculated as follows: \[ p_{e} = (9.11 \times 10^{-31} \ \mathrm{kg})(3.4 \times 10^{6} \ \mathrm{m/s}) = 3.10 \times 10^{-24} \ \mathrm{kg \cdot m/s} \]
Similarly, for a proton with a mass of \( 1.67 \times 10^{-27} \ \mathrm{kg} \), moving at the same velocity, the momentum is \[ p_{p} = (1.67 \times 10^{-27} \ \mathrm{kg})(3.4 \times 10^{6} \ \mathrm{m/s}) = 5.68 \times 10^{-21} \ \mathrm{kg \cdot m/s} \]
From these calculations, it's clear that due to their different masses, the electron and proton have different momenta even at the same velocity. Momentum is therefore essential in determining a particle's wavelength via de Broglie's equation.

Kinetic Energy

Kinetic Energy (K.E.) is the energy of motion and is given by the equation \[ K = \frac{1}{2} mv^2 \]
This equation can be rearranged to solve for the velocity (v) if the kinetic energy and mass (m) are known, which is useful in the latter part of the exercise. The rearranged form is: \[ v = \sqrt{ \frac{2K}{m}} \]
For example, if an electron has a kinetic energy of \( 2.7 \times 10^{-15} \ \mathrm{J} \), its velocity can be calculated as follows: \[ v_{e} = \sqrt{ \frac{2 \times 2.7 \times 10^{-15}}{9.11 \times 10^{-31}}} = 2.45 \times 10^7 \ \mathrm{m/s} \]
For a proton with the same kinetic energy, the velocity would be: \[ v_{p} = \sqrt{ \frac{2 \times 2.7 \times 10^{-15}}{1.67 \times 10^{-27}}} = 1.80 \times 10^6 \ \mathrm{m/s} \]
These calculations show that for the same kinetic energy, particles with different masses will have different velocities, affecting their momenta and consequently their wavelengths.
Kinetic energy is a vital piece of the puzzle, connecting the energies of particles to their quantum mechanical properties.

Particle Wavelength

The concept of 'particle wavelength' bridges the gap between wave and particle descriptions of matter. From de Broglie's equation, \[ \lambda = \frac{h}{p} \]
where \( \lambda \) is the wavelength and \( p \) is the momentum, we can see that every particle has an associated wavelength. In our exercise, we calculated the wavelengths for both an electron and a proton given their momenta.
For instance, the wavelength of an electron with a momentum of \( 3.10 \times 10^{-24} \ \mathrm{kg \cdot m/s} \) is: \[ \lambda_{e} = \frac{h}{p_{e}} = \frac{6.626 \times 10^{-34}}{3.10 \times 10^{-24}} = 2.14 \times 10^{-10} \ \mathrm{m} \]
In contrast, a proton with a momentum of \( 5.68 \times 10^{-21} \ \mathrm{kg \cdot m/s} \) has a wavelength: \[ \lambda_{p} = \frac{h}{p_{p}} = \frac{6.626 \times 10^{-34}}{5.68 \times 10^{-21}} = 1.17 \times 10^{-13} \ \mathrm{m} \]
These results illustrate that due to their larger mass and thus smaller momentum for a given speed, protons have much shorter wavelengths than electrons.
This understanding is vital to explain various phenomena in quantum mechanics and supports the wave-particle duality nature of matter.