An \(\alpha\) -particle is accelerated from rest through a potential difference of \(6.0 \mathrm{~V}\). Its de-Broglie wavelength is (a) \(5 \AA\) (b) \(4.15 \mathrm{pm}\) (c) \(414.6 \AA\) (d) \(5 \mathrm{~nm}\)

Understanding how an alpha particle accelerates is crucial in physics, especially in the context of nuclear science and quantum mechanics. An alpha particle, symbolized as \(\alpha\), is essentially a helium nucleus comprising two protons and two neutrons, which means it carries a +2 charge due to the protons. When an alpha particle is subjected to an electric field, such as the potential difference mentioned in the exercise, it experiences a force proportional to the electric field and its charge.

Now, recall that force is directly related to acceleration as per Newton's second law (F = ma), where 'm' is mass and 'a' is acceleration. Because the alpha particle starts at rest, any acceleration will increase its velocity from zero to a certain value, depending on the potential difference it traverses. This change in velocity translates to an increase in kinetic energy, which can then be used to perform various calculations, including the determination of the de-Broglie wavelength.

Kinetic energy (KE) represents the energy possessed by an object due to its motion. It's a fundamental concept in both classical and quantum physics, determining how energy transforms and transfers through different states. The formula for kinetic energy, \(\frac{1}{2}mv^2\), where 'm' is mass and 'v' is velocity, shows that it's directly proportional to the mass and the square of the velocity of the moving object.

In the context of charged particles like alpha particles, their kinetic energy after acceleration through a potential difference derives from the work done by the electric force. As the solution outlines, we calculate it by multiplying the charge by the potential difference. This is a critical step in eventually finding out the de-Broglie wavelength because it gives us the energy conversion necessary to solve for the particle’s momentum.

The momentum of an object is a measure of its motion and is given as the product of its mass and velocity (\(p = mv\)). In classical mechanics, it's a vector quantity, directly related to the amount of force and time over which it's applied. However, diving into quantum mechanics introduces momentum as also a crucial part of wave-particle duality characterized by the de-Broglie hypothesis.

For the alpha particle example, solving for momentum is a stepping stone toward finding its de-Broglie wavelength. By equating the kinetic energy to \(\frac{1}{2}mv^2\) and since momentum is \(mv\), we can rearrange to find momentum as \(p = \sqrt{2mKE}\). Importantly, the mass of the alpha particle is the sum of the masses of its constituent protons and neutrons, which emphasizes the relationship between mass and momentum in atomic-scale particles.

Planck's constant, denoted as \(h\), is a fundamental constant in physics that relates the energy of a photon to its frequency: \(E = hu\). The value of \(h\) (\(6.626 \times 10^{-34} J\cdot s\)) has profound implications in quantum mechanics, being central to the quantum of action that underpins the theory's axioms. Its introduction led to the dawning of the quantum era, signifying a discrete rather than a continuous universe at the smallest scales.

In our problem-solving context, Planck's constant helps us calculate the de-Broglie wavelength of our alpha particle. Using the equation \(\lambda = \frac{h}{p}\), we connect the concept of momentum with wave-like properties of matter, with \(h\) as the proportionality factor that makes this relationship computable. Thus, Planck’s constant not only bridges the gap between matter and energy but also provides a pathway to calculate physical properties like wavelength from an object's momentum.