Question

Find the ratios of de Broglie wavelengths of a \(100-\mathrm{MeV}\) proton to a 100 - MeV electron.

Short answer

Question: Calculate the ratio of de Broglie wavelengths of a 100 MeV proton to a 100 MeV electron. Answer: To calculate the ratio of de Broglie wavelengths, first find the momenta of both the proton and the electron using their given energy and mass. Then, use the de Broglie wavelength formula, \(\lambda = \dfrac{h}{p}\), to find the wavelengths for both particles. Finally, divide the wavelength of the proton by the wavelength of the electron to find the ratio, \(\dfrac{\lambda_p}{\lambda_e} = \dfrac{p_e}{p_p}\).

Step-by-step solution

Recall the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \(\lambda = \dfrac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle.

Relate energy and momentum for both particles

The total energy of a particle, considering relativistic effects, is given by: \(E = \sqrt{(pc)^2 + (mc^2)^2}\), where E is the energy, p is the momentum, m is the mass of the particle, and c is the speed of light. In this exercise, both particles have the same energy of 100 MeV. We can use this to find momentum for both particles. For electron, \(E_e = 100\ \text{MeV}\) and \(m_e = 9.11 \times 10^{-31}\ \text{kg}\). For proton, \(E_p = 100\ \text{MeV}\) and \(m_p = 1.67 \times 10^{-27}\ \text{kg}\).

Calculate momentum for both particles

We need to find the momentum of both particles using their masses and energies. Rearrange the energy formula to get: \(p=\sqrt{(E^2-(mc^2)^2)}/c\) Calculate momentum for the electron: \(p_e =\sqrt{(100 \times 10^6eV \times 1.6 \times 10^{-19} J/eV)^2 - ((9.11\times10^{-31}kg)(3\times10^8m/s)^2)^2}/(3\times10^8m/s)\) Calculate momentum for the proton: \(p_p =\sqrt{(100 \times 10^6eV \times 1.6 \times 10^{-19} J/eV)^2 - ((1.67\times10^{-27}kg)(3\times10^8m/s)^2)^2}/(3\times10^8m/s)\)

Calculate de Broglie wavelengths for both particles

Use the de Broglie wavelength formula to find the wavelengths: \(\lambda_e = \dfrac{h}{p_e}\) and \(\lambda_p = \dfrac{h}{p_p}\), where \(h = 6.626 \times 10^{-34}\ {\rm Js}\).

Find the ratio of wavelengths

Finally, find the ratio of de Broglie wavelengths of a proton to an electron: \(\dfrac{\lambda_p}{\lambda_e} = \dfrac{\dfrac{h}{p_p}}{\dfrac{h}{p_e}} = \dfrac{p_e}{p_p}\) Substitute the values of \(p_e\) and \(p_p\) to find the final ratio. This will give us the desired ratio of de Broglie wavelengths of a 100 MeV proton to a 100 MeV electron.

Key concepts

Planck's constant

Planck's constant, represented by the symbol \( h \), is a fundamental parameter in quantum mechanics. It has a value of approximately \( 6.626 \times 10^{-34} \text{ Js} \) (joule seconds) and it is the quantum of electromagnetic action that relates the energy carried by a photon to its frequency. More formally, Planck's constant defines the relationship between energy \( E \) and the frequency \( u \) of a photon: \[ E = hu \].

This constant is also essential in explaining the particle-wave duality concept, which is exemplified in the de Broglie hypothesis. According to de Broglie, every particle exhibits wave-like properties, with the de Broglie wavelength \( \text{(}\text{λ}\text{)} \) of the particle being inversely proportional to its momentum \( p \), as defined by the formula \( \text{λ} = \frac{h}{p} \). Understanding Planck's constant offers insight into the quantized nature of light and matter, providing a bridge between classical and quantum theories.

Particle momentum

In classical mechanics, momentum of a particle is defined as the product of its mass \( m \) and velocity \( v \), represented by \( p = mv \). However, in the context of quantum mechanics and relativity, the concept of momentum is more nuanced. For particles with mass, momentum becomes more significant at high velocities (near the speed of light, \( c \)), where relativistic effects must be considered.

The de Broglie wavelength's dependence on momentum means that as the momentum of a particle increases, its wavelength decreases. This inverse relationship is crucial in understanding the behaviour of particles at atomic and subatomic scales. Since particles at these scales can have very high velocities, their momentum cannot be accurately described using classical mechanics alone, which leads to the inclusion of relativistic modifications for high energy particles, such as those encountered in particle accelerators or cosmic rays.

Relativistic energy-momentum relationship

The relativistic energy-momentum relationship is a cornerstone of Einstein's theory of relativity, enriching our understanding of how energy, momentum, and mass relate at high velocities. It is given by the equation \[ E = \text{└}\!\!\!└ sqrt{(pc)^2 + (mc^2)^2} \], where \( E \) is the total energy of the particle, \( p \) is its momentum, \( m \) its rest mass, and \( c \) the speed of light in vacuum. It shows that the energy of a particle grows rapidly as its velocity approaches the speed of light, reflecting the necessity to include the energy contribution of mass as well as the kinetic energy.

When dealing with high-energy particles like protons and electrons (as in the given textbook problem), the energies are significant enough that relativistic effects cannot be ignored. Hence, finding the particle momentum is not as straightforward as in classical physics, and we must use this relationship to compute momentum correctly for high-speed particles. This calculation is crucial for predicting observed phenomena, such as particle speeds in particle accelerators or the decay products of high-energy collisions in particle physics experiments.