Explanation:
It is given that,
The ionization energy of a nitrogen atom, [tex]E=2.32\ aJ[/tex]
aJ is attpjoules. [tex]1\ aJ=10^{-18}\ J[/tex]
[tex]E=2.32\times 10^{-18}\ J[/tex]
The energy of an atom is given by :
[tex]E=h\times f[/tex]
f is the frequency of the photons
[tex]f=\dfrac{E}{h}[/tex]
[tex]f=\dfrac{2.32\times 10^{-18}}{6.67\times 10^{-34}}[/tex]
[tex]f=3.47\times 10^{15}\ Hz[/tex]
Also, [tex]E=\dfrac{hc}{\lambda}[/tex]
[tex]\lambda=\dfrac{hc}{E}[/tex]
[tex]\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{2.32\times 10^{-18}}[/tex]
[tex]\lambda=8.62\times 10^{-8}\ m[/tex]
Hence, this is the required solution.
