Question

Photo-dissociation of water $H_{2}O(1)+hv\rightarrow H_{2}(g)+\dfrac{1}{2}O_{2}(g)$has been suggested as a source of hydrogen. The heat absorbed in this reaction is $289.5kJ/mol$ of water decomposed. The maximum wavelength that would provide the necessary energy assuming that one photon causes the dissociation of water molecule is $(1eV=96.5kJ/mol)$
\[\begin{align}
  & A.6.95\times {{10}^{-28}}m \\
 & B.4.19\times {{10}^{-7}}m \\
 & C.6.95\times {{10}^{-31}}m \\
 & D.1.72\times {{10}^{-6}}m \\
\end{align}\]

Answer 1

Hint: Here, the energy of the light produces the dissociation of the water molecules as given in the equation. Then, from the given energy we can calculate the wavelength of the light which produced the reaction.

Formula used: $E=hv$and $c=v\lambda$

Complete step-by-step solution:
We know from photoelectric effect, that when an incident beam of light falls on a metal, it radiates or releases electrons. The energy of these electrons depends on the frequency of the light, and in turn, depends on the wavelength of the light source.
We know that $E=hv$, where $E$ is the energy of the electron which is produced due to light rays of frequency $v$ and $h$ is the Planck’s constant. We also know that $c=v\lambda$, where $c$ is the speed of light whose frequency is $v$ and wavelength $\lambda$
Here, given that the heat absorbed by $1\; mol$ of $H_{2}O$ is $289.5kJ/mol$.
Then, $1\;molecule$ of $H_{2}O$ absorbs,$\dfrac{289.5kJ/mol}{6.022\times 10^{23}}=4.81\times 10^{-19}J$
Then, the wavelength which produces $4.81\times 10^{-19}J=\dfrac{hc}{\lambda}.$
Substituting the values, we get, $\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{4.81\times 10^{-19}} =\lambda.$
$\implies \lambda=4.19\times 10^{-7}m$
Thus the answer is \[B. 4.19\times {{10}^{-7}}m\]
Additional information:
From, photoelectric effect, we know that the intensity of the electrons emitted depends on the wavelength. But however, it is independent of the intensity of the light waves. Also there is almost no time lag between the light incident on the metal and the electron being emitted.

Note: The photoelectric effect talks of light incident on a metal. Here we have a light incident on water, which produces photons. Since both the processes are similar, we can use the same formula and logic to solve them. Also, not that the given energy is for one mole of water, thus we need to calculate the energy required per molecule to solve the sum.