Assuming that the smallest measurable wavelength in an experiment is 0.410 fm (femtometers), what is the maximum mass of an object traveling at 815 m*s^-1 for which the de Broglie wavelength is observable?
w = wavelength
w = h/mv
Convert w to meters, solve for m in kg.
5.03*10-4
@DrBob22
I think you would be of more help if you actually answered the question instead of just giving the equation in all your posts.
To find the maximum mass of an object for which the de Broglie wavelength is observable, we need to use the formula for de Broglie wavelength:
λ = h / p
where λ is the wavelength, h is Planck's constant (h = 6.626 x 10^-34 J*s), and p is the momentum of the object.
First, let's rearrange the formula to solve for momentum:
p = h / λ
Given the smallest measurable wavelength (λ) of 0.410 fm (femtometers) and the speed (v) of 815 m/s, we can calculate the momentum:
p = h / λ
= (6.626 x 10^-34 J*s) / (0.410 x 10^-15 m)
= 1.615 x 10^-19 kg*m/s
Now, we can find the maximum mass (m) by rearranging the formula for momentum:
p = mv
Rearranging the formula:
m = p / v
Substituting the values:
m = (1.615 x 10^-19 kg*m/s) / (815 m/s)
= 1.982 x 10^-22 kg
Therefore, the maximum mass of an object for which the de Broglie wavelength is observable, given the conditions, is approximately 1.982 x 10^-22 kg.