Assuming that the smallest measurable wavelength in an experiment is 0.410 fm (femtometers), what is the maximum mass of an object traveling at 395 m·s–1 for which the de Broglie wavelength is observable?
wavelength = h/mv
You know wavelength (convert to m), you know h, and you know velocity. Solve for mass (in kg).
To solve this problem, we need to use the de Broglie wavelength formula, which relates the wavelength of a particle to its mass and velocity.
The de Broglie wavelength (λ) can be expressed as:
λ = h / p
where h is the Planck's constant (6.626 x 10^-34 Js) and p is the momentum of the particle.
The momentum of a particle can be found using the formula:
p = m * v
where m is the mass of the particle and v is its velocity.
Given that the smallest measurable wavelength is 0.410 fm (femtometers), we can convert it to meters by dividing by 10^15:
λ = 0.410 fm / (10^15 m/fm) = 0.410 x 10^-15 m
Now, we can rearrange the de Broglie wavelength formula to solve for mass:
m = h / (λ * v)
Substituting the values, we have:
m = (6.626 x 10^-34 Js) / (0.410 x 10^-15 m * 395 m/s)
Simplifying:
m = (6.626 x 10^-34 Js) / (1.6175 x 10^-13 Js)
m ≈ 4.083 x 10^-21 kg
Therefore, the maximum mass of an object traveling at 395 m/s for which the de Broglie wavelength is observable is approximately 4.083 x 10^-21 kg.