# A photon having lambda = 854*10^(-10) m causes the ionization of a nitrogen atom. Give the I.E. per mole of nitrogen in KJ?

Jul 13, 2017

${\text{IE" = "1402.78068 kJ mol}}^{- 1}$

#### Explanation:

When a photon hits the electron in the outer shell, it excites the electron proportionally according to the photon's frequency, and the electron is ejected when it surpasses the orbital potential energy.

Now you need to calculate the energy of the photon.

$E = \frac{h c}{\lambda}$,

(assuming that the light is in vacuum)

where $h$ is Planck's Constant in Joules seconds which is equivalent to $6.63 \times {10}^{-} 34 \text{J"cdot"s}$

$\text{c}$ is speed of light which is equal to $3 \times {10}^{8} \text{m/s}$. If you want the closest value because this is rounded off a quick search can be of help.

$\lambda$ is the wavelength of the photon in $m$, but as the value is already in $m$, it does not need to be converted.

$E = \left(6.63 \times {10}^{-} 34 \text{J" cdot cancel"s" xx 3xx10^8cancel"m""/"cancel"s")/( 854*10^(-10) cancel"m}\right)$

$E = 2.3290398 \times {10}^{-} 18$ $\text{J}$

According to the question this is the energy required for ionization of atom of nitrogen so this can be represented as

$\text{IE" = (2.3290398xx10^-18"J")/"atom}$

therefore (2.3290398xx10^-18"J")/cancel"atom" * (6.023 xx 10^23 cancel"atom")/"mole"

$\text{IE""/mole" = ("1402780.68 J")/"mole}$

$\text{1 kJ}$ $=$ $\text{1000 J}$

$\implies \text{IE""/mol" = (1402780.68 cancel"J" * "kJ"/(1000cancel"J"))/"mol}$

${\text{IE" = "1402.78068 kJ mol}}^{- 1}$