Question

Suppose we produce X-rays not by smashing elections into targets but by smashing protons, which are far more massive. If the same accelerating potential difference were used for both, how would the cutoff wavelengths of the two X-ray spectra compare? Explain.

Short answer

The cutoff wavelengths of the x-ray spectra produced by protons would be shorter than the x-ray spectra produced by electrons. This is because protons have larger mass compared to electrons.

Step-by-step solution

Recall the de Broglie wavelength formula and the definition of kinetic energy

In this case, we'll be using the de Broglie wavelength formula \( \lambda = \frac{h}{p} \) where \( h \) is the Planck's constant and \( p \) is the momentum. Remember also that kinetic energy is given by \( KE = \frac{1}{2} mv^2 \) where \( m \) is mass and \( v \) is velocity.

Relate kinetic energy, momentum and wavelength

Since both protons and electrons have the same kinetic energy under the same accelerating potential, we can equate their kinetic energies to get: \( \frac{1}{2}m_ev_e^2 = \frac{1}{2}m_pv_p^2 \). Solving this for \( v_p \) gives \( v_p = v_e \sqrt{ \frac{m_e}{m_p}} \). Substituting this into the de Broglie wavelength formula for protons gives: \( \lambda_p = \frac{h}{m_p v_p} = \frac{h}{m_p v_e \sqrt{ \frac{m_e}{m_p}}} = \frac{h}{v_e \sqrt{m_e^3}} \). And for electrons, we know that \( \lambda_e = \frac{h}{m_e v_e} \).

Find the ratio of the cutoff wavelengths

We can now find the ratio of the cutoff wavelength of protons (\( \lambda_p \)) to that of electrons (\( \lambda_e \)): \( \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{v_e \sqrt{m_e^3}}}{\frac{h}{m_e v_e}} = \sqrt{m_e} \). Remember that the mass of an electron (\( m_e \)) is much less than the mass of a proton (\( m_p \)), so this ratio is less than one. This means that the cutoff wavelength for protons is less than that for electrons, under the same accelerating potential difference.

Key concepts

de Broglie Wavelength

The de Broglie wavelength is a concept that reveals the wave-particle duality of matter. The idea is that every particle, regardless of its size, has a wavelength associated with it. This can be calculated using the formula \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. Momentum, in simple terms, is the product of an object's mass and velocity.
This concept tells us that even particles as small as electrons have wave-like properties. Understanding the de Broglie wavelength is crucial when exploring atomic and subatomic scales, where classical physics doesn't always give an accurate description.
For example, when protons and electrons are accelerated, their respective de Broglie wavelengths will differ due to their different masses, even if they have the same kinetic energy. This results in different physical properties, like cutoff wavelengths, which are vital in fields like quantum mechanics and material science.

Kinetic Energy

Kinetic energy is the energy that an object possesses because of its motion. When particles such as protons or electrons are accelerated, they gain kinetic energy, which can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( KE \) is the kinetic energy, \( m \) is the mass of the particle, and \( v \) is its velocity.
In the context of producing X-rays, both electrons and protons can be given the same amount of kinetic energy by accelerating them through the same potential difference. Their velocities may vary due to their differing masses, but their energy levels remain equivalent under these conditions.
This concept helps us understand why smaller, lighter particles like electrons will move faster than massive particles like protons when subjected to the same accelerating potential, impacting their wavelengths and subsequent X-ray production.

Cutoff Wavelength

The cutoff wavelength is a concept that arises in the context of X-ray production. It is the shortest wavelength possible for a set of emitted X-rays and is directly tied to the maximum energy of the incoming particles. The equation \( \lambda = \frac{hc}{eV} \) relates the cutoff wavelength \( \lambda \) with Planck's constant \( h \), the speed of light \( c \), the elementary charge \( e \), and the accelerating potential \( V \).
When particles collide with an atomic target, they produce X-rays with a range of wavelengths. The cutoff wavelength represents the highest energy, shortest wavelength, X-ray emitted.
When comparing mediums such as protons and electrons, despite experiencing the same accelerating potential, heavier protons have longer wavelengths than lighter electrons. Therefore, the cutoff wavelength for electron-induced X-rays is shorter, reflecting their lighter mass and higher velocity when both are accelerated under the same conditions.

Accelerating Potential Difference

The accelerating potential difference refers to the voltage difference through which particles are accelerated. It provides energy to the particles, transforming their potential energy into kinetic energy. When charged particles like electrons and protons are accelerated through a voltage difference, their velocity increases. This is crucial in experiments designed to produce X-rays.
The accelerating potential is directly related to the kinetic energy formula \( KE = eV \), where \( KE \) is the kinetic energy, and \( e \) is the charge of the electron.
Consequently, particles with higher kinetic energy, due to higher potential differences, will generate X-rays of higher energy and shorter wavelength. An equivalent accelerating voltage results in protons and electrons with the same kinetic energy, but this affects their momentum and de Broglie wavelength differently due to mass differences. This is why, when using the same accelerating potential, lighter electrons will yield X-rays with shorter cutoff wavelengths than heavier protons.