Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

To solve the problem of calculating the momentum and the de Broglie wavelength of electrons that have been accelerated through a potential difference of 56 volts, we need to apply some fundamental physics principles. Let's break it down into two parts: momentum and de Broglie wavelength.

Calculating Momentum

When electrons are accelerated through a potential difference, they gain kinetic energy equal to the work done on them by the electric field. The kinetic energy (KE) gained by the electrons can be expressed as:

KE = eV

Here, e is the charge of an electron (approximately \(1.6 \times 10^{-19}\) coulombs), and V is the potential difference (56 V in this case). Therefore, the kinetic energy can be calculated as follows:

KE = (1.6 \times 10^{-19} \text{ C})(56 \text{ V}) = 8.96 \times 10^{-18} \text{ J}

The kinetic energy is also related to the momentum (p) of the electron by the equation:

KE = \frac{p^2}{2m}

Where m is the mass of the electron, approximately \(9.11 \times 10^{-31}\) kg. Rearranging this equation allows us to solve for momentum:

p = \sqrt{2m \cdot KE}

Plugging in the values:

• m = 9.11 \times 10^{-31} \text{ kg}
• KE = 8.96 \times 10^{-18} \text{ J}

p = \sqrt{2 \cdot (9.11 \times 10^{-31}) \cdot (8.96 \times 10^{-18})}

Calculating this gives:

p \approx 4.22 \times 10^{-24} \text{ kg m/s}

Determining de Broglie Wavelength

Next, let's find the de Broglie wavelength of the electrons. The de Broglie wavelength (\(\lambda\)) is given by the formula:

\(\lambda = \frac{h}{p}\)

Where h is Planck's constant, approximately \(6.63 \times 10^{-34} \text{ J s}\). We have already calculated momentum (p), so we can substitute this into the equation:

\(\lambda = \frac{6.63 \times 10^{-34}}{4.22 \times 10^{-24}}\)

Calculating this gives:

\(\lambda \approx 1.57 \times 10^{-10} \text{ m}\)

Summary of Results

In summary, for electrons accelerated through a potential difference of 56 V:

• Momentum (p): \(4.22 \times 10^{-24} \text{ kg m/s}\)
• de Broglie Wavelength (λ): \(1.57 \times 10^{-10} \text{ m}\)

These calculations highlight the wave-particle duality of electrons, showcasing how they exhibit both particle-like and wave-like properties depending on the context. Understanding these concepts is fundamental in the field of quantum mechanics and helps us appreciate the behavior of subatomic particles.