Question

Certain sun glasses having small crystals of $AgCl$ incorporated in the lenses, on exposure to light of appropriate wavelength turns to grey colour to reduce the glare following the reaction:
$AgCl\stackrel{hv}{\to }\underset{\left(Grey\right)}{Ag}+Cl$
If the heat of reaction for the decomposition of $AgCl$ is $248kJmo{l}^{-1}$, what maximum wavelength is needed to induce the desired process?

• A. $4.83\times {10}^{-4}m$
• B. $9.6\times {10}^{-7}m$
• C. $6\times {10}^{-7}m$
• D. $4.83\times {10}^{-7}m$

Answer 1

The correct option is D $4.83\times {10}^{-7}m$

The heat of reaction for the decomposition of $AgCl$ is $248kJmo{l}^{-1}$, that is $248000Jmo{l}^{-1}$. If we convert this $Jmo{l}^{-1}$ to $Jphoto{n}^{-1}$, we get $\frac{248000}{6.022\ast {10}^{23}}$ $Jphoto{n}^{-1}$, that is $4.12\ast {10}^{-19}$$Jphoto{n}^{-1}$.

The conversion factor being $1mole=$$6.022\ast {10}^{23}photons$.

Thus, we can say that the heat of reaction for the decomposition of $AgCl$ is $4.12\ast {10}^{-19}$$Jphoto{n}^{-1}$.

We also know the formula $E=$$\frac{hc}{\lambda }$, where $E$ is the energy which is $4.12\ast {10}^{-19}$$Jphoto{n}^{-1}$, $h$ is the Planck's constant, $c$ is the velocity of light in vacuum and $\lambda$ is the wavelength of light.

Putting all the values in the equation we get,-

$4.12\ast {10}^{-19}$$=$$\frac{6.62\ast {10}^{-34}\times 3\times {10}^8}{\lambda }$

Thus $\lambda$$=$$\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{4.12\times {10}^{-19}}$, that is $4.82\times {10}^{-7}m$

Thus the required wavelength is $4.82\times {10}^{-7}m$

Option $D$ is the correct answer.