Answer
2.0 eV, or 45 kcal/mole.
Work Step by Step
The photon energy equals the energy required for the chemical reaction. Use equation 27–4, E = hf, to find the energy of a photon. For electromagnetic radiation, we also know that $f=c/\lambda$.
$$E=hf=\frac{hc}{\lambda}=\frac{1240 eV\cdot nm}{630nm}=2.0eV$$
This is the required energy per molecule to initiate the reaction. Convert eV/molecule to kcal/mol.
$$E= \frac{1.968eV}{molecule}\frac{1.60\times10^{-19}J}{eV}\frac{kcal}{4186J}\frac{6.02\times10^{23}molecules}{mole}$$
$$=45kcal/mole$$
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Another solution:
Since the film is sensitive to light, so the energy needed for a chemical reaction is given by
$$ E=hf$$
where $h$ is Planck’s constant and $f$ is the frequency of the photon which is given by $c=f\lambda$.
Thus,
$$E=\dfrac{hc}{\lambda} $$
Plugging the known;
$$ E=\dfrac{6.36\times10^{-34}\times 3\times10^8}{630\times10^{-9}} =\bf 3\times10^{-19}\;\rm J$$
The author needs the answer in $\rm eV$, so
$$ E= \rm 3\times10^{-19}\;\rm J\times \dfrac{1\;eV}{1.6\times10^{-19}\;J}=\color{red}{\bf 1.893}\;\rm eV$$
The author also needs the answer in $\rm kcal/mol$, so
$$ E= \rm \dfrac{3\times10^{-19}\;\rm J}{1\;molecule}\times \dfrac{1\;kcal}{4186\;J}\times\dfrac{6.02\times10^{23}\;molecule}{1\;mol}$$
$$E=\color{red}{\bf 43.1}\;\rm kcal/mol$$
