The de Broglie wavelength of a particle can be calculated using the following formula:
λ = h / p
Where: λ is the de Broglie wavelength, h is the Planck's constant (approximately 6.626 x 10^-34 joule-seconds), p is the momentum of the particle.
To determine the momentum of an electron accelerated through a potential difference of 1 keV, we need to use the relation between the kinetic energy (KE) and momentum (p) of a particle:
KE = (1/2)mv^2 = p^2 / (2m)
Where: m is the mass of the electron (approximately 9.11 x 10^-31 kg), v is the velocity of the electron.
Given that the electron is accelerated from rest, its initial kinetic energy is zero. Therefore, the kinetic energy gained through the potential difference is equal to the 1 keV:
KE = eV
Where: e is the elementary charge (approximately 1.6 x 10^-19 C), V is the potential difference (1 keV = 1.6 x 10^-16 J).
Setting the kinetic energy equal to the potential difference:
eV = p^2 / (2m)
Solving for momentum (p):
p = √(2m eV)
Now we can substitute this value of momentum into the de Broglie wavelength formula:
λ = h / √(2m eV)
Plugging in the known values:
λ ≈ 6.626 x 10^-34 J s / √(2 * 9.11 x 10^-31 kg * 1.6 x 10^-19 C * 1.6 x 10^-16 J)
Evaluating the expression:
λ ≈ 6.626 x 10^-34 J s / √(5.82 x 10^-60 J^2 s^2 C kg)
Finally, simplifying the expression:
λ ≈ 6.626 x 10^-34 J s / (2.41 x 10^-30 J s C kg)
Therefore, the de Broglie wavelength of an electron accelerated from rest through a potential difference of 1 keV is approximately:
λ ≈ 2.75 x 10^-5 meters