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Question: The light sensitive compound on most photographic films is silver bromide AgBr. A film is exposed wh...

The light sensitive compound on most photographic films is silver bromide AgBr. A film is exposed when the light energy absorbed dissociates this molecule into atoms. The energy of dissociation of AgBr is 105 J⋅mol−1{10^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}s. For a photon that is just able to dissociate a molecule of AgBr, the photon energy is
A. 1.04 eV1.04\,{\text{eV}}
B. 2.08 eV2.08\,{\text{eV}}
C. 3.12 eV3.12\,{\text{eV}}
D. 4.16 eV4.16\,{\text{eV}}

Explanation

Solution

Remind the concept of Avogadro’s number. Relate this concept with the present problem to determine that total number of molecules of silver bromide that will have the given dissociation energy. Then determine the dissociation energy of one silver bromide molecule and convert it into electronvolt.

Complete step by step answer:
The dissociation energy E1 mol{E_{1\,{\text{mol}}}} for one mole of silver bromide is given as 105 J⋅mol−1{10^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}.
E1 mol=105 J⋅mol−1{E_{1\,{\text{mol}}}} = {10^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}
We have to determine the energy that the photon should possess to dissociate only one silver bromide molecule.
We know that one mole of any compound contains the number of molecules of that compound equal to
the Avogadro’s number.
The value of the Avogadro’s number is 6.023×10236.023 \times {10^{23}}.
NA=6.023×1023{N_A} = 6.023 \times {10^{23}}
Hence, one mole of silver bromide will contain 6.023×10236.023 \times {10^{23}} molecules of the silver bromide.
Therefore, the energy 105 J⋅mol−1{10^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}} is the dissociation energy for 6.023×10236.023 \times {10^{23}} molecules of silver bromide.
We have to determine the energy EE of only one molecule of silver bromide.
Divide the dissociation energy E1 mol{E_{1\,{\text{mol}}}} of one mole of silver bromide by the Avogadro’s number NA{N_A} to determine the dissociation energy EE of one silver bromide molecule.
E=E1 molNAE = \dfrac{{{E_{1\,{\text{mol}}}}}}{{{N_A}}}
Substitute 105 J⋅mol−1{10^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}} for E1 mol{E_{1\,{\text{mol}}}} and 6.023×1023 molecules6.023 \times {10^{23}}\,{\text{molecules}} for NA{N_A} in the above equation.
E=105 J⋅mol−16.023×1023 moleculesE = \dfrac{{{{10}^5}\,{\text{J}} \cdot {\text{mo}}{{\text{l}}^{ - 1}}}}{{6.023 \times {{10}^{23}}\,{\text{molecules}}}}
⇒E=1.66×10−19 J\Rightarrow E = 1.66 \times {10^{ - 19}}\,{\text{J}}
Hence, the energy of dissociation for one molecule of silver bromide is 1.66×10−19 J1.66 \times {10^{ - 19}}\,{\text{J}}.
Converting this energy is dissociation from joule to electron volt.
⇒E=(1.66×10−19 J)(1 eV1.6×10−19 J)\Rightarrow E = \left( {1.66 \times {{10}^{ - 19}}\,{\text{J}}} \right)\left( {\dfrac{{1\,{\text{eV}}}}{{1.6 \times {{10}^{ - 19}}\,{\text{J}}}}} \right)
⇒E=1.04 eV\Rightarrow E = 1.04\,{\text{eV}}

Therefore, the energy that a photon should have to dissociate one molecule of silver bromide is 1.04 eV1.04\,{\text{eV}}.

Hence, the correct option is A.

Note:
The students should not forget that the energy obtained by dividing the given dissociation energy of one mole of silver bromide by Avogadro’s number is in joule as all the physical quantities used have the units in SI system of units. All the options given have the energy in the unit electronvolt. Hence, the obtained energy value should be converted into electron vo to get the correct answer.