Question

it takes 155. kj/mol to break a fluorine - fluorine single bond. calculate the maximum wavelength of light for which a fluorine - fluorine single bond could be broken by absorbing a single photon.
be sure your answer has the correct number of significant digits.

nm

Explanation:

Step1: Calculate the energy per photon

The energy required to break one \(F - F\) bond is \(E=\frac{155\times10^{3}\text{ J/mol}}{6.022\times 10^{23}\text{ mol}^{-1}}\)

$$E=\frac{155000}{6.022\times 10^{23}}\text{ J}\approx2.574\times 10^{-19}\text{ J}$$

Step2: Use the formula \(E = h

u=\frac{hc}{\lambda}\) to find \(\lambda\)
We know that \(h = 6.626\times 10^{-34}\text{ J}\cdot\text{s}\) and \(c=3\times 10^{8}\text{ m/s}\)
From \(\lambda=\frac{hc}{E}\), substitute the values:
\(\lambda=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{2.574\times 10^{-19}}\text{ m}\)

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Convert meters to nanometers: \(1\text{ m}=10^{9}\text{ nm}\), so \(\lambda = 7.72\times 10^{-7}\times10^{9}\text{ nm}=772\text{ nm}\)

Answer:

\(772\text{ nm}\)