Assuming that the smallest measurable wavelength in an experiment is 0.990 fm (femtometers), what is the maximum mass of an object traveling at 269 m·s–1 for which the de Broglie wavelength is observable?
See your post below. Convert fm to m.
To find the maximum mass for which the de Broglie wavelength is observable, we can use the de Broglie wavelength formula:
λ = h / p,
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the object.
In this case, we have the smallest measurable wavelength (λ) of 0.990 fm (femtometers) and the velocity (v) of the object, which allows us to calculate the momentum (p) using the equation:
p = mv,
where m is the mass of the object and v is the velocity.
First, let's convert the smallest measurable wavelength to meters:
0.990 fm = 0.990 x 10^-15 m.
Now, we can rearrange the de Broglie wavelength formula to solve for the mass (m):
m = h / (λv).
Substituting the values into the formula:
m = (6.626 x 10^-34 J·s) / ((0.990 x 10^-15 m) * (269 m·s^-1)).
m = 2.389 x 10^-20 kg.
Therefore, the maximum mass of an object traveling at a velocity of 269 m·s^-1 for which the de Broglie wavelength is observable is approximately 2.389 x 10^-20 kilograms.