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The first ionization energy, E , of a nitrogen atom is 2.32 aJ. What is the wavelength of light, in nanometers, that is just sufficient to ionize a nitrogen atom? Values for constants can be found in the Chempendix.

Answer :

Answer:

Required wavelength of light is 85.6 nm.

Explanation:

E = h×ν = [tex]\frac{hc}{\lambda}[/tex]

E = Energy in J

ν = frequency (Hz s-1)

λ = wavelength in m  

Planck constant h = J s

c = speed of light = m s-1  

2.32 aJ = J

E = h×ν  

[tex]\lambda = \frac{2.998\times10^{8}}{3.50\times10^{15}}=8.56\times10^{-8} m = 85.6 nm[/tex]

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