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Chapter 13: Problem 15

Photo-dissociation of water \(\mathrm{H}_{2} \mathrm{O}(\mathrm{I})+h v\) \(\rightarrow \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) has been suggested as a source of hydrogen. The heat absorbed in this reaction is \(289.5 \mathrm{~kJ} / \mathrm{mole}\) of water decomposed. The maximum wavelength that would provide the necessary energy assuming that one photon causes the dissociation of one water molecule is \((1 \mathrm{eV}=96.5 \mathrm{~kJ} / \mathrm{mol})\) (a) \(413.33 \mathrm{~nm}\) (b) \(826.67 \mathrm{~nm}\) (c) \(206.67 \mathrm{~nm}\) (d) \(4.3 \mathrm{~nm}\)

Short Answer

Expert verified
The maximum wavelength that would provide the necessary energy is about 413.33 nm, so the correct option is (a) 413.33 nm.

Step by step solution

01

Convert energy required per mole to energy per photon

First, convert the heat absorbed from kJ/mole to Joules per mole. 1 kJ = 1000 J. Then use the energy of one mole of photons (Avogadro's number) to find the energy per photon.
02

Use the Energy-Photon Wavelength Equation

Use the equation that relates the energy of a photon to its wavelength, which is given by the equation E = hc/位, where h is Planck's constant, c is the speed of light, and 位 is the wavelength of the photon.
03

Solve for Maximum Wavelength

Rearrange the equation to solve for 位 and substitute the values for h, c, and the energy per photon calculated in Step 1 to find the maximum wavelength.
04

Check the Possible Answers

Check the calculated wavelength against the multiple-choice options to find the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Per Photon
The concept of energy per photon is crucial in understanding photo-dissociation and many other photochemical reactions. In physics, a photon is a particle representing a quantum of light or other electromagnetic radiation. The energy of a single photon is determined by its frequency (u), which is related to the speed of light (c) and its wavelength (lambda) through the equation c = lambda u.

When we talk about the energy per photon in the context of photo-dissociation of water, we refer to the individual energy packets that photons must carry to break the chemical bonds in water molecules. This energy can be calculated by using the equation E = h u, where h is Planck's constant, and u is the frequency of the photon. However, given that photon frequency and wavelength are inversely proportional (the higher the frequency, the shorter the wavelength), this relationship can also be shown as E = hc/lambda.

Understanding energy per photon is vital because it determines whether a photon has sufficient energy to cause photo-dissociation of water molecules. If the energy is too low, the photon cannot break the bonds, and no reaction will occur.
Photon Wavelength Equation
The photon wavelength equation lies at the heart of understanding how light interacts with matter. It is expressed as E = hc/lambda, where E is the energy of the photon, h is Planck's constant, c is the speed of light in a vacuum, and lambda is the wavelength. This equation effectively bridges the gap between the classical wave description of light and its quantum particle perspective.

In the exercise, we use this foundational equation to calculate the maximum wavelength of light required to induce photo-dissociation in water molecules. As the wavelength lambda increases, the energy E of each photon decreases, so there is a maximum wavelength that can still produce the necessary energy to trigger the reaction. Solving for lambda in the equation, we get lambda = hc/E, and we can see that for a larger lambda, we have less energy per photon, based on the fact that both h and c are constants.

Using this equation, students can better understand how high-energy photons with shorter wavelengths can break chemical bonds, while photons with wavelengths longer than the maximum will not have enough energy to cause photo-dissociation.
Planck's Constant
Planck's constant, denoted as h, is a fundamental constant in quantum mechanics. It has a value of approximately 6.626 x 10^-34 joules-seconds. Planck's constant relates the energy of a photon to its frequency, as shown in the equation E = hu. It represents the proportionality constant between the minimum energy increments that can be exchanged and the frequency of the electromagnetic wave.

In our exercise involving photo-dissociation, Planck's constant serves as the bridge between the quantum world of photons and the macroscopic world we can measure, for example in the energy required to break the bonds in water molecules. The immutable nature of Planck's constant allows us to perform precise calculations when dealing with photon energies, frequencies, and wavelengths. It's fundamental to our understanding of quantum phenomena and plays a decisive role when calculating the maximum wavelength capable of inducing the photo-dissociation of water.

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Most popular questions from this chapter

The dye acriflavine when dissolved in water has its maximum light absorption at \(4530 \AA\) and has maximum florescence emission at \(5080 \AA\). The number of fluorescence quanta is about \(53 \%\) of the number of quanta absorbed. What percentage of absorbed light energy is emitted as fluorescence? (a) \(41 \%\) (b) \(47 \%\) (c) \(74 \%\) (d) \(63 \%\)

The vapours of \(\mathrm{Hg}\) absorb some electron accelerated by a potential difference of \(5.0 \mathrm{~V}\) as a result of which light is emitted. If the full energy of single incident electron is supposed to be converted into light emitted by single Hg-atom, the wavelength of the emitted light is (a) \(2480 \mathrm{~nm}\) (b) \(248 \mathrm{~nm}\) (c) \(6200 \mathrm{~nm}\) (d) \(620 \mathrm{~nm}\)

Orbital with maximum symmetry is (a) p-orbital (b) s-orbital (c) \(d_{x y}\) -orbital (d) \(d_{z^{2}}\) -orbital

If \(\lambda_{0}\) is the threshold wavelength for photoelectric emission from a metal surface, \(\lambda\) is the wavelength of light falling on the surface of metal and \(m\) is the mass of electron, then the maximum speed of ejected electrons is given by (a) \(\left[\frac{2 h}{m}\left(\lambda_{0}-\lambda\right)\right]^{1 / 2}\) (b) \(\left[\frac{2 h c}{m}\left(\lambda_{0}-\lambda\right)\right]^{1 / 2}\) (c) \(\left[\frac{2 h c}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda_{0} \lambda}\right)\right]^{1 / 2}\) (d) \(\left[\frac{2 h}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)\right]^{1 / 2}\)

Light of wavelength, \(\lambda\), strikes a metal surface with intensity \(X\) and the metal emits \(Y\) electrons per second of maximum kinetic energy \(Z\). What will happen to \(Y\) and \(Z\) if \(X\) is halved? (a) \(Y\) will be halved and \(Z\) will be doubled (b) \(Y\) will be doubled and \(Z\) will be halved (c) \(Y\) will be halved and \(Z\) will remain the same (d) \(Y\) will remain same and \(Z\) will be halved
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