Final answer:
Using the de Broglie wavelength formula, the maximum mass of an object traveling at 339 m/s to have an observable de Broglie wavelength of 0.790 fm is approximately 2.50 x 10^-24 kg.
Step-by-step explanation:
To calculate the maximum mass of an object for which the de Broglie wavelength can be observed, given the smallest measurable wavelength and the object's speed, we use the de Broglie wavelength formula: \(\lambda = \frac{h}{mv}\), where \(\lambda\) is the wavelength, \(h\) is Planck's constant (\(6.626 \times 10^{-34} Js\)), \(m\) is the mass, and \(v\) is the velocity. If the smallest measurable wavelength is \(0.790 fm\) (\(0.790 \times 10^{-15} m\)) and the object is traveling at \(339 ms^{-1}\), the equation is rearranged to solve for mass \(m\):
\(m = \frac{h}{\lambda v}\)
Plugging in the known values:
\(m = \frac{6.626 \times 10^{-34} Js}{(0.790 \times 10^{-15} m)(339 ms^{-1})}\)
By calculating the above expression, we find the maximum mass that meets the condition for an observable de Broglie wavelength:
\(m \approx 2.50 \times 10^{-24} kg\)
Therefore, an object with a mass of up to approximately \(2.50 \times 10^{-24} kg\), moving at \(339 ms^{-1}\), would have a de Broglie wavelength of \(0.790 fm\), which is the limit of observability in this scenario.