Introduction:
The de Broglie wavelength is a concept in quantum mechanics that describes the wave-like behavior of particles. It is given by the equation λ = h/p, where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.

Given:
Potential difference (V) = 1000 eV

Calculating the momentum:
The potential difference can be converted into energy using the equation E = qV, where E is the energy, q is the charge of the particle, and V is the potential difference.

Since the electron has a charge of -1.6 x 10^-19 C, we can calculate the energy as follows:
E = (-1.6 x 10^-19 C)(1000 eV)
E = -1.6 x 10^-19 J

The momentum of the electron can be calculated using the equation p = √(2mE), where p is the momentum, m is the mass of the electron, and E is the energy.

The mass of the electron is approximately 9.11 x 10^-31 kg, so we can calculate the momentum as follows:
p = √(2(9.11 x 10^-31 kg)(-1.6 x 10^-19 J))
p ≈ 1.30 x 10^-24 kg·m/s

Calculating the de Broglie wavelength:
Now that we have the momentum of the electron, we can calculate its de Broglie wavelength using the equation λ = h/p.

The Planck's constant is approximately 6.63 x 10^-34 J·s, so we can calculate the de Broglie wavelength as follows:
λ = (6.63 x 10^-34 J·s)/(1.30 x 10^-24 kg·m/s)
λ ≈ 5.10 x 10^-11 m

Explanation:
When an electron is accelerated by applying a potential difference, it gains energy. This energy can be converted into momentum using the equation p = √(2mE), where p is the momentum, m is the mass of the electron, and E is the energy.

The de Broglie wavelength of the electron can then be calculated using the equation λ = h/p, where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the electron.

In this case, applying a potential difference of 1000 eV to the electron results in an energy of -1.6 x 10^-19 J. The momentum of the electron is then calculated to be approximately 1.30 x 10^-24 kg·m/s. Finally, using the de Broglie wavelength equation, the wavelength is calculated to be approximately 5.10 x 10^-11 m.

Therefore, the de Broglie wavelength associated with the electron accelerated by a potential difference of 1000 eV is approximately 5.10 x 10^-11 m.