Question

A proton is fired from very far away towards a nucleus with charge \(Q=120 e\), where \(e\) is the electronic charge. It makes a closest approach of \(10 \mathrm{fm}\) to the nucleus. The de Broglie wavelength (in units of \(\mathrm{fm}\) ) of the proton at its start is: (take the proton mass, \(m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ;\) \(\left.\mathrm{h} / e=4.2 \times 10^{-15} \mathrm{~J} . \mathrm{s} / \mathrm{C} ; \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)\)

Short answer

The de Broglie wavelength of the proton at its start is calculated using the potential energy at the closest approach and the expression for the de Broglie wavelength.

Step-by-step solution

Calculate the potential energy at the closest approach

Firstly, calculate the potential energy (PE) of the proton when it is at its closest approach of 10 fm from the nucleus using Coulomb's law. The formula for potential energy in an electric field is: PE = k * Q * q / r, where k is Coulomb's constant, Q is the nucleus charge, q is the charge of the proton, and r is the distance of closest approach.

Find the initial kinetic energy

The proton starts from very far away, where its potential energy due to the nuclear force is zero. All potential energy at the closest approach is converted into kinetic energy due to the conservation of energy. So, the initial kinetic energy (KE) of the proton is equal to the potential energy at the closest approach.

Determine the proton's initial momentum

The kinetic energy of a particle is given by the equation KE = p^2 / (2 * m), where p is momentum and m is the mass of the proton. Use this equation to find the proton's initial momentum.

Calculate the de Broglie wavelength of the proton

The de Broglie wavelength \( \lambda \) is given by the equation \( \lambda = h / p \), where h is Planck's constant and p is the momentum of the proton. Use the momentum from step 3 and h/e (also given), where you need to multiply by the electronic charge to get Planck's constant in J*s, to calculate the de Broglie wavelength.

Express the de Broglie wavelength in femtometers

The value of the wavelength will be initially in meters. To convert this to femtometers (fm), use the conversion that \(1 \mathrm{fm} = 10^{-15} \mathrm{~m}\).

Key concepts

Coulomb's law

Coulomb's law is a fundamental principle in physics that describes the force between two charged objects. According to this law, the electrostatic force (\f\( F \f\)) acting between two point charges is directly proportional to the product of the magnitudes of the charges (\f\( q_1 \f\) and \f\( q_2 \f\)) and inversely proportional to the square of the distance (\f\( r \f\)) between them.

\f[ F = k \frac{q_1 q_2}{r^2} \f]
Here, \f\( k \f\) is Coulomb’s constant, equal to \f\( 8.9875 \times 10^9 \f\) Nm²/C². In the context of the exercise, Coulomb's law allows us to calculate the electrostatic potential energy between a proton and a nuclear charge. This lays the foundation for understanding how forces at the atomic level work and is crucial when considering the interactions between a nucleus with charge \f\( Q=120e \f\) and a proton.

Kinetic energy to potential energy conversion

In classical mechanics, energy conversion between kinetic (energy of motion) and potential (stored energy) forms is a pivotal concept. This energy transformation is observed when a proton is fired towards a nucleus. The kinetic energy, which the proton has initially due to its motion, is gradually converted into electrostatic potential energy as it gets closer to the nucleus, up to the point of closest approach.

Because of energy conservation, the kinetic energy (\f\( KE \f\)) of the proton, at a very far distance (where potential energy is negligible), will be equal to the potential energy (\f\( PE \f\)) at the closest approach point, disregarding any energy loss:
\f[ KE_{initial} = PE_{closest} = k \frac{Q q}{r} \f]
Understanding this energy conversion is critical for interpreting how the proton behaves as it approaches the nucleus and also in calculating the de Broglie wavelength of the proton.

Momentum and de Broglie wavelength relationship

The relationship between momentum and de Broglie wavelength is one of the cornerstones of quantum mechanics. According to de Broglie, particles such as electrons or protons have wave-like properties, and their wavelength (\f\( \f\)) can be described by the equation:

\f[ \f = \frac{h}{p} \f]
Where \f\( h \f\) is Planck's constant and \f\( p \f\) is the momentum of the particle. This relationship implies that the wavelength is inversely proportional to the momentum: as the momentum increases, the wavelength decreases, and vice versa. In the context of the exercise, once the initial kinetic energy and hence the momentum of the proton is known, the de Broglie wavelength can be calculated, providing insight into the particle’s wave-like behavior at the quantum scale.

Planck's constant application

Planck's constant (\f\( h \f\)), a fundamental constant in quantum mechanics, is used to relate the energy of a photon to its frequency. Its value is approximately \f\( 6.62607015 \times 10^{-34} \f\) joule seconds (Js).

In the given problem, Planck's constant is used in its relationship to momentum (\f\( p \f\)) to find the de Broglie wavelength (\f\( \f\)) of the proton:
\f[ \f = \frac{h}{p} \f]
This application is crucial for determining the wavelength associated with particles, illustrating the quantum nature of matter, and is a key component in calculating the de Broglie wavelength for the proton in this exercise.

Nuclear charge and proton interaction

The interaction between a nuclear charge and a proton is a manifestation of the electromagnetic force, one of the four fundamental forces in physics. The magnitude of the nuclear charge (\f\( Q \f\)), especially when it is that of multiple protons as in this exercise, has a significant impact on the electrostatic potential energy and consequently on the proton's behavior. As the proton moves towards the positively charged nucleus, their charges interact according to Coulomb's law, where the repulsive force and the associated potential energy increase as the distance decreases.

This interaction is critical in not only providing the potential energy at the closest approach but also in facilitating the calculations required to derive the proton's de Broglie wavelength. The example exercise involves a substantial nuclear charge (\f\( Q=120e \f\)), leading to a quantifiable and significant potential energy at the closest approach of 10 femtometers (\f\( fm \f\)), which in turn is used to resolve the de Broglie wavelength of the proton.