An electron of mass m is moving in a uniform electric field
E⃗ = −2E0 î
where E0 is a positive constant. Initially, the electron has velocity
V⃗ = V0 î
and its initial de-Broglie wavelength is
λ0 = h/(4mv0)
Find the de-Broglie wavelength of the electron at time t.
The de-Broglie wavelength of a particle is given by:
λ = h/p
where p is the momentum of the particle.
Since the particle is an electron, its charge is −e. The electric field is directed along the negative x-axis:
E = −2E0 î
The force on the electron is:
F = qE
= (−e)(−2E0)
= 2eE0
Thus, the electron experiences a constant force in the positive x-direction.
Applying Newton's second law:
ma = 2eE0
Hence:
a = 2eE0/m
Using the first equation of motion:
v = v0 + at
v = v0 + (2eE0t)/m
The momentum at time t becomes:
p = mv
= mv0 + 2eE0t
The initial wavelength is:
λ0 = h/(4mv0)
Therefore:
h = 4mv0λ0
Substituting into the de-Broglie relation:
λ = h/(mv0 + 2eE0t)
= 4mv0λ0 /(mv0 + 2eE0t)
Dividing numerator and denominator by mv0:
λ = 4λ0 /(1 + 2eE0t/mv0)
As time increases, the momentum of the electron increases due to acceleration by the electric field. Since wavelength is inversely proportional to momentum, the de-Broglie wavelength continuously decreases with time.
λ = 4λ0 /(1 + 2eE0t/mv0)