Question

the first ionization energy, e, of a boron atom is 8.01 x 10² kj/mol. what is the wavelength, in nm, that is just sufficient to ionize a boron atom? report your answer using 3 significant figures.

Explanation:

Step1: Calculate the energy per atom

We know that \(1\space mol = 6.022\times10^{23}\space atoms\).
The energy per atom \(E\) (in \(J\)) is given by:
\(E=\frac{8.01\times 10^{2}\times10^{3}\space J/mol}{6.022\times 10^{23}\space atoms/mol}\)
\(E=\frac{8.01\times10^{5}}{6.022\times10^{23}}\space J\)
\(E = 1.33\times10^{-18}\space J\)

Step2: Use the Planck - Einstein relation

The Planck - Einstein relation is \(E = h
u=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant) and \(c=3\times 10^{8}\space m/s\) (speed of light).
We can solve for \(\lambda\):
\(\lambda=\frac{hc}{E}\)
Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 3\times10^{8}\space m/s\) and \(E=1.33\times 10^{-18}\space J\)
\(\lambda=\frac{6.626\times10^{-34}\times3\times 10^{8}}{1.33\times10^{-18}}\space m\)
\(\lambda=\frac{19.878\times10^{-26}}{1.33\times10^{-18}}\space m\)
\(\lambda = 1.49\times10^{-7}\space m\)

Step3: Convert meters to nanometers

Since \(1\space m=10^{9}\space nm\)
\(\lambda=1.49\times10^{-7}\times10^{9}\space nm\)
\(\lambda = 149\space nm\)

Answer:

\(149\space nm\)