Question

general chemistry 4th edition mcquarrie·rock·gallogly university science books presented by macmillan learning assuming that the smallest measurable wavelength in an experiment is 0.890 fm, what is the maximum mass of an object traveling at 437 m·s⁻¹ for which the de broglie wavelength is observable? m = kg tools ×10ʸ

Explanation:

Step1: Convert the wavelength unit

The de - Broglie wavelength formula is \(\lambda=\frac{h}{mv}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant), \(\lambda\) is the wavelength, \(m\) is the mass, and \(v\) is the velocity.
Given \(\lambda=0.890\space fm = 0.890\times10^{-15}\space m\), \(v = 437\space m/s\)

Step2: Rearrange the de - Broglie formula to solve for \(m\)

From \(\lambda=\frac{h}{mv}\), we can solve for \(m\) as \(m=\frac{h}{\lambda v}\)

Substitute the values:
\(m=\frac{6.626\times 10^{-34}\space J\cdot s}{(0.890\times 10^{-15}\space m)\times(437\space m/s)}\)

First, calculate the denominator: \((0.890\times 10^{-15})\times(437)=0.890\times437\times10^{-15}=388.93\times10^{-15}\)

Then, \(m=\frac{6.626\times 10^{-34}}{388.93\times 10^{-15}}\)

Using the rule of exponents \(\frac{a\times10^{n}}{b\times10^{m}}=\frac{a}{b}\times10^{n - m}\), we have \(m=\frac{6.626}{388.93}\times10^{-34 + 15}\)

\(\frac{6.626}{388.93}\approx0.017\)

\(m = 0.017\times10^{-19}=1.7\times10^{-21}\space kg\)

Answer:

\(1.7\times10^{-21}\)