Question

Certain sunglasses having small crystals of $AgCl$ incorporated in the lenses, on exposure to light of appropriate wavelength turns to grey color to reduce the glare following the reaction:
$AgCl\xrightarrow{{h\upsilon }}A{g_{\left( {grey} \right)}} + Cl$ .
If the heat of reaction for the decomposition of $AgCl$ is $248kJ/mol$ , 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

Hint:
The amount of energy given is in terms of kilojoules. We will first convert the amount of energy from kilojoules to joules and from joules to number of moles of photon using Avogadro’s number.

Complete step by step answer:
The energy of a photon is defined as the amount of energy required to travel by photon.
The energy of a photon is directly proportional to the speed of light and inversely proportional to wavelength.
It is given by the formula as follows:
$E = \dfrac{{hc}}{\lambda }$
where, $E = $ energy of photon
$h = $ Planck’s constant
$c = $ speed of light
$\lambda = $ wavelength.
Energy of photon in terms of frequency is given as follows:
where, $E = $ energy of photon
$h = $ Planck’s constant
$f = $ frequency.
Given:
$AgCl\xrightarrow{{h\upsilon }}A{g_{\left( {grey} \right)}} + Cl$
Amount of energy$\left( E \right) = 248kJ/mol$
We will convert kilojoules to using the following conversion
$1kJ = 1000Joules$
Therefore,
$ \Rightarrow 248kJ/mol = 248 \times 1000$
$ \Rightarrow 248000J/moles$
Now we will convert $J/mol$to $J/photon$ using Avogadro’s Number as follows:
$1mol = 6.023 \times {10^{23}}$ photons.
We will divide amount of energy in terms of joule by Avogadro’s number we get,
$ \Rightarrow \dfrac{{248000}}{{6.023 \times {{10}^{23}}}}$
$ = 4.12 \times {10^{ - 19}}J/photon$ .
Now we have the value of amount of energy in terms of photons.
Using these data we will calculate the wavelength needed to induce the process.
Given data:
$E = 4.12 \times {10^{ - 19}}J/photon$
$h = 6.62 \times {10^{ - 34}}J.s$
$c = 3 \times {10^8}m/s$
To find: $\lambda = ?$
Formula to be used: $E = \dfrac{{hc}}{\lambda }$