Physics – De Broglie Wavelenght Counting?

Brilliant Queen (BQ)_forever !!!: Question in Physics.
1. If the De Broglie wavelenght of an electron is equal to 5.00 * 10 ^ -7, how fast is the electron moving?
2. What is the De Broglie wavelenght of 1375 Kg car traveling at 43 km/h?

Answers and Views:

Answer by victeric
The de Broglie wavelength is given by λ*p = h, where λ is the de Broglie wavelength, p is momentum, and h is Planck’s constant. If we’re going to do these calculations relativistically, we need to use p = γ*m*v, where γ = [1-(v/c)^2]^(-1/2), where c is the speed of light. If you don’t need the relativistic correction, just use p = m*v.

1). p = h/λ = γ*m*v = m*v*[1-(v/c)^2]^(-1/2). You’re given λ, and h and c are constants, so you can solve for v. If p = m*v, this is just h/λ = m*v => v = h/(m*λ).

2). λ = h/p. The car is traveling much slower than the speed of light, so just use p = m*v, so that λ = h/(m*v). You’re given m and v, and h is a constant, so you can calculate λ.

Answer by schmiso
The first de Broglie equation relates the wavelength λ to the particle momentum p as:
λ = h/p = h/(γ·m·v)
h is Planck’s constant, m mass of the particle, v speed
γ = sqrt(1- (v/c)²)

If the particles is much lower than speed of light, Lorenz factor is close to unity:
v< γ≈1
and the relation simplifies to
λ = h/p = h/(m·v)

1.)
λ = h/(m·v)

v = h/(m·λ)
= 6.6260693×10^-34 Js / (9.109 3826×10^-31kg · 5.0×10^-7m)
= 1454m/s

2.)
v = 43km/h = 11.94m/s

λ = h/(m·v)
= 6.6260693×10^-34 Js / (1375kg · 11.94m/s)
= 4.034×10^-38