The light sensitive compound on photographic films is silver bromide AgBr- A film is exposed when the light energy absorbed dissociates this molecule into its atoms- The energy of dissociation of AgBr is 10-5 J mol-1- For a photon that is just able to dissociate a molecule of AgBr- the photon energy is1-04 eV2-08 eV3-12 eV4-16 eV

Question

Toppr Admin · Accepted Answer

1 mole of AgBr contains 6-022-xD7-1023 molecules and 1J of energy is equivalent to 6-24-xD7-1018eV-Thus- energy needed to dissociate 1 molecule of AgBr is given by-E-105-xD7-6-24-xD7-10186-022-xD7-1023-1-04eV

1 mole of AgBr contains 6.022×1023 molecules and 1J of energy is equivalent to 6.24×1018eV.Thus, energy needed to dissociate 1 molecule of AgBr is given by:E=105×6.24×10186.022×1023=1.04eV

1 mole of AgBr contains $6.022 \times 10^{23}$ molecules and $1 J$ of energy is equivalent to $6.24 \times 10^{18} e V$ .
Thus, energy needed to dissociate 1 molecule of AgBr is given by:

$E = \frac{10^{5} \times 6.24 \times 10^{18}}{6.022 \times 10^{23}} = 1.04 e V$