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When discussing de Broglie wavelength, a crucial step is converting kinetic energy from various units to a common one that fits our calculations. In the provided exercise, the kinetic energy is originally measured in MeV (mega-electron volts), a common unit in atomic physics. However, to proceed with calculations, it must be converted to Joules (J). This transformation is essential because the formulas we use in physics, like those for energy and momentum, typically require Joules as the standard unit.
Here's how the conversion is done:

• 1 electron volt (eV) equals approximately \(1.602 \times 10^{-19}\) Joules.
• 1 MeV is \(10^6\) times an eV, which equals \(1.602 \times 10^{-13}\) Joules.

For instance, if you have a proton with kinetic energy of 2.0 MeV:- Multiply 2.0 MeV by the conversion factor \(1.602 \times 10^{-13} \) J/MeV to get \(3.204 \times 10^{-13}\) Joules.Similarly, a kinetic energy of 10.0 MeV converts to \(1.602 \times 10^{-12}\) Joules using the same method.
Understanding these conversions is essential in physics, as energy in different forms can usually interconvert, allowing deeper insights into how systems behave energetically.

In physics, momentum is a measure of the motion of an object, and it plays a significant role when calculating the de Broglie wavelength of a particle. Here, we'll guide you through the link between kinetic energy and momentum using a fundamental relationship. Kinetic energy (E) is related to momentum (p) and mass (m) using the equation: \[ E = \frac{p^2}{2m} \].This equation can be rearranged to solve for momentum:- \( p = \sqrt{2Em} \).Turn the mathematical equation into practical application by considering that:

• The mass of a proton (\(m\)) is approximately \(1.67 \times 10^{-27}\) kilograms.

For the proton with kinetic energy of 2.0 MeV (converted to \(3.204 \times 10^{-13}\) Joules):- The momentum calculation is \(p_{2.0 \text{ MeV}} = \sqrt{2 \times 3.204 \times 10^{-13} \times 1.67 \times 10^{-27}}\), resulting in \(4.429 \times 10^{-20}\) kg m/s.Similarly, for the 10.0 MeV case (converted to \(1.602 \times 10^{-12}\) Joules):- The momentum is \(p_{10.0 \text{ MeV}} = \sqrt{2 \times 1.602 \times 10^{-12} \times 1.67 \times 10^{-27}}\), resulting in \(9.893 \times 10^{-20}\) kg m/s.These steps underline the process of moving from kinetic energy to momentum, which sets the stage for further calculation of the de Broglie wavelength.

Planck's constant is a fundamental element in quantum mechanics that influences our understanding of particles' wave-like behavior. When exploring the de Broglie wavelength, Planck's constant (denoted as \(h\)) acts as a bridge between momentum and wavelength, showcasing the wave-particle duality of matter. The value of Planck's constant is approximately \(6.626 \times 10^{-34}\) Joule seconds (Js), a very small value that highlights the scale at which quantum effects become significant. To calculate the de Broglie wavelength \(\lambda\), you use the formula:- \( \lambda = \frac{h}{p} \), where \(p\) is the momentum.This equation reveals that the wavelength is inversely proportional to momentum; as the momentum increases, the wavelength decreases, and vice versa. This helps to visualize how even particles like protons have wave-like characteristics.
For example, using momentum calculations:

• For 2.0 MeV, with \(p = 4.429 \times 10^{-20} \) kg m/s, the de Broglie wavelength is approximately \( \lambda = \frac{6.626 \times 10^{-34}}{4.429 \times 10^{-20}} = 1.497 \times 10^{-14}\) m.
• For 10.0 MeV, with \(p = 9.893 \times 10^{-20} \) kg m/s, the wavelength is approximately \( \lambda = \frac{6.626 \times 10^{-34}}{9.893 \times 10^{-20}} = 6.700 \times 10^{-15}\) m.

This calculation illustrates the fascinating principle that all matter presents wave-like properties, instrumental in quantum mechanics.