91Ó°ÊÓ

The bond energy of H-I

$\mathbf{E}\left(\mathrm{HI}\right)\mathbf{=}\mathbf{295}\mathbf{kJ}\mathbf{/}\mathbf{mol}$

To find out the longest wavelength that can dissociate a molecule of HI, we calculate the energy of one molecule of HI

$\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}\mathbf{=}\frac{E(H-I)}{{\mathbf{N}}_A}$

Where ${\mathbf{N}}_A$is Avogadro’s number. value of ${\mathbf{N}}_A$ is$\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{×}{\mathbf{10}}^{23}\mathbf{molecules}\mathbf{/}\mathbf{mol}$

So,

$\begin{array}{c}\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}\mathbf{=}\frac{\mathbf{295}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{×}{\mathbf{10}}^3\mathbf{J}\mathbf{/}\mathbf{kJ}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{×}{\mathbf{10}}^{23}\mathbf{molecules}\mathbf{/}\mathbf{mol}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}\mathbf{/}\mathbf{molecule}\end{array}$

Therefore, the maximum wavelength of the HI molecule can be calculated as

$\mathbf{λ}\mathbf{=}\frac{\mathrm{hc}}{\mathbf{E}\mathbf{(}\mathbf{HI}{\mathbf{)}}_{\mathrm{molecule}}}$

Where h is Planck’s constant. Value of h is$\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{sec}$ . And c is the speed of light. Value of c is $\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{sec}$.

$\begin{array}{c}\mathbf{λ}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{×}\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}\mathbf{.}\mathbf{Mol}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{05}\mathbf{×}{\mathbf{10}}^{-7}\mathbf{m}\\ \mathbf{=}\mathbf{405}\mathbf{.}\mathbf{78}\mathbf{nm}\end{array}$

Thus, the longest wavelength that can dissociate a molecule oh HI is 405.78nm

Given wavelength = 254nm

First of all, find the energy corresponding to this wavelength and then subtract the bond dissociation energy of HI.

So, the energy corresponding to this wavelength

$\begin{array}{l}\mathbf{E}\mathbf{=}\frac{\mathrm{hc}}{\mathrm{λ}}\text{ }\\ \text{ â¶Ä„â¶Ä„}\mathbf{=}\frac{\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{×}{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{×}\mathbf{3}\mathbf{×}{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{254}\mathbf{×}{\mathbf{10}}^{-9}\mathbf{m}}\end{array}$

Excess energy = energy corresponding to 254nm wavelength- bond dissociation

The energy of a molecule oh HI

$\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{82}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{8987}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}\\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}\end{array}$

Thus, the excess energy over that needed for dissociation is $\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{J}$.

If the excess energy is carried away by the H atom as kinetic energy, the speed is calculated by the formula $\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$. where m is the H atom's mass. The value of m is$\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{×}{\mathbf{10}}^{-24}\mathbf{g}$. And v is the speed.

Therefore,

$\mathbf{KE}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$

$\begin{array}{l}\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{×}{\mathbf{10}}^{-19}\mathbf{=}\frac{1}{2}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{×}{\mathbf{10}}^{-24}\mathbf{×}{\mathbf{v}}^2\\ {\mathbf{v}}^2\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{92}\mathbf{×}{\mathbf{10}}^{-19}}{\mathbf{8}\mathbf{.}\mathbf{365}\mathbf{×}{\mathbf{10}}^{-24}}\\ {\mathbf{v}}^2\text{ â¶Ä„â¶Ä„}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{349}\mathbf{×}{\mathbf{10}}^{-5}\\ \text{ }\mathbf{v}\text{ â¶Ä„}\mathbf{=}\text{ }\mathbf{590}\mathbf{m}\mathbf{/}\mathbf{s}\end{array}$

The speed is 590 m/s if the extra energy is transferred to the H atom as kinetic energy.