Question

The first ionization energy, E, of a nitrogen atom is 2.32 aJ. What is the wavelength of light, in nm, that is just sufficient to ionize a nitrogen atom?

Answer 1

Use the formula: \(E_{photon}=\dfrac{hc}{\lambda}\)

where c is the speed of light, h plancks constant and \(\lambda\) the wavelength.

Remember to use SI units.

Answer 2

Ok, so I get the fact that:
\[Ephoton=\frac{ (6.626 x 10^{-34}J*s)(2.9979*10^{8}m*s^{-1}) }{? }\]
What I still don't understand is what wavelength is & where does the first ionization of nitrogen plays a role in this problem?

Answer 3

The energy the photon needs to be is given in the question, 2.32 aJ.

\(\sf 1~attojoule=1.0*10^{-18}J\)

so, \(\sf E_{photon}=2.32~aJ=2.32*10^{-18}J\)

wavelength is what you are solving for, they're asking for you to express it in nm.

\(1 ~nm=1.0*10^{-9}~m\)