A photon having $\lambda =800{{A}^{0}}$ causes the ionization of a nitrogen atom. The I.E. per mole of nitrogen in kJ is?
[Given: E = 1240eV-nm, 1 eV = $96.43kJ/mol$]

Hint: The basic concept of the relation between the wavelength of the light with the energy and the frequency that is the energy of a photon equation given by $E=\dfrac{hc}{\lambda }$ will lead you to the required answer.

Complete step by step answer:
- In the physical chemistry chapters, we have studied about the various parameters and derivations like calculation of Wavelength of light using the De – Broglie equation, calculation of energy of a photon and many other entities.
- Let us now calculate the ionization energy of the nitrogen atom based on these concepts.
- Photon is the elementary particle which is defined as the small packet of energy.
- To find the energy of this photon the very common equation which relates wavelength with the frequency and energy is used.
- This equation is given by,
\[E=h\nu \]
where h is the Planck’s constant and it has the value of $6.625\times {{10}^{-34}}Js$
\[\nu \] is the frequency of the photon.
The above equation can be written as,
$E=\dfrac{hc}{\lambda }$ where, $\nu =\dfrac{c}{\lambda }$
and c is the velocity of light which has the value of $3\times {{10}^{8}}m/s$
$\lambda $ is the wavelength of light given.
- Now according to the data $\lambda = 800{{A}^{0}}$
By substituting all these values in the above equation, we get
\[E=\dfrac{6.625\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m/s}{800\times {{10}^{-10}}m}\] [Since,$1{{A}^{0}} = {{10}^{-10}}m$]
\[\Rightarrow E = 2.47\times {{10}^{-18}}J = 2.47\times {{10}^{-21}}kJ\]
- Now, let us calculate the energy associated with one mole of the photon that is of nitrogen is,
\[E = 6.022\times {{10}^{23}}\times 2.47\times {{10}^{-21}}kJ\]
\[\Rightarrow E = 1495.42kJ/mol\]
Therefore, the correct answer is ionization energy per mole of nitrogen \[1495.42kJ/mol\]

Note: Note that according to the Avogadro law, one mole of any substance contains Avogadro's number of molecules that is \[6.022\times {{10}^{23}}\] number of molecules and therefore the conversion of the value into per mole of the substance involves the multiplication of this number.