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Chapter 7: Problem 54

X rays have wavelengths on the order of \(1 \times 10^{-10} \mathrm{m} .\) Calcu- late the energy of \(1.0 \times 10^{-10} \mathrm{m} \mathrm{X}\) rays in units of kilojoules per mole of \(\mathrm{X}\) rays. AM radio waves have wavelengths on the order of \(1 \times 10^{4} \mathrm{m}\) . Calculate the energy of \(1.0 \times 10^{4} \mathrm{m}\) radio waves in units of kilojoules per mole of radio waves. Consider that the bond energy of a carbon-carbon single bond found in organic compounds is 347 \(\mathrm{kJ} / \mathrm{mol}\) . Would \(\mathrm{x}\) rays and/or radio waves be able to disrupt organic compounds by breaking carbon-carbon single bonds?

Short Answer

Expert verified
The energy of X rays is \(119.4 \mathrm{kJ/mol}\), which is higher than the bond energy of carbon-carbon single bond (347 \(\mathrm{kJ/mol}\)), so X rays can disrupt organic compounds by breaking carbon-carbon single bonds. The energy of radio waves is \(1.195 \times 10^{-4} \mathrm{kJ/mol}\), which is much lower than the bond energy, so radio waves cannot break carbon-carbon single bonds.

Step by step solution

01

Identifying the given values

The given values are: - Wavelength of X rays: \(\lambda_{X}\) = \(1.0 \times 10^{-10} m\) - Wavelength of radio waves: \(\lambda_{radio}\) = \(1.0 \times 10^{4} m\) - Bond energy of carbon-carbon single bond: 347 \(kJ/mol\) - Calculate the energy of X rays and radio waves in \(kJ/mol\), and compare them to the bond energy of carbon-carbon single bond.
02

Calculate the energy of the X rays

To calculate the energy of the X rays, we can use the Planck's formula: \[E = hf = \frac{hc}{\lambda}\] where - E is the energy of the X rays - h is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{J s}\)) - c is the speed of light (\(3.0 \times 10^{8} \mathrm{m/s}\)) - f is the frequency of the X rays - \(\lambda\) is the wavelength of the X rays Substitute the given values into the Planck's formula: \(E_{X} = \frac{(6.626 \times 10^{-34} \mathrm{J s})(3.0 \times 10^{8} \mathrm{m/s})}{1.0 \times 10^{-10} \mathrm{m}}\)
03

Convert energy of X rays to \(kJ/mol\)

To convert the energy of X rays to kilojoules per mole, we can use the conversion factors: - Multiply the energy by \(10^3\) to convert it from joules to kilojoules. - Multiply the energy by Avogadro's number (\(6.022 \times 10^{23} \mathrm{mol}^{-1}\)) to convert it from energy per photon to energy per mole. \(E_{X} = (1.986 \times 10^{-25} \mathrm{J}) (10^3) (6.022 \times 10^{23} \mathrm{kJ/mol})\)
04

Calculate the energy of the radio waves

Just as in step 2, we can now calculate the energy \(E_{radio}\) of the radio waves using the Planck's formula: \(E_{radio} = \frac{(6.626 \times 10^{-34} \mathrm{J s})(3.0 \times 10^{8} \mathrm{m/s})}{1.0 \times 10^{4} \mathrm{m}}\)
05

Convert energy of radio waves to \(kJ/mol\)

Similar to step 3, we convert the energy of radio waves to kilojoules per mole: \(E_{radio} = (1.986 \times 10^{-29} \mathrm{J}) (10^3) (6.022 \times 10^{23} \mathrm{kJ/mol})\)
06

Compare the energies with the bond energy of carbon-carbon single bond

At this step, we compare the energy of X rays and radio waves in \(kJ/mol\) with the bond energy of a carbon-carbon single bond (347 \(kJ/mol\)): - Energy of X rays: \(E_{X} = 119.4 \mathrm{kJ/mol}\) - Energy of radio waves: \(E_{radio} = 1.195 \times 10^{-4} \mathrm{kJ/mol}\) Since the energy of X rays is higher than the bond energy of the carbon-carbon single bond, they can potentially disrupt organic compounds by breaking carbon-carbon single bonds. On the other hand, the energy of radio waves is much lower than the bond energy, so they cannot cause any breakage of the carbon-carbon single bonds.

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Most popular questions from this chapter

How many unpaired electrons are present in each of the following in the ground state: $\mathrm{O}, \mathrm{O}^{+}, \mathrm{O}^{-}, \mathrm{Os}, \mathrm{Zr}, \mathrm{S}, \mathrm{F}, \mathrm{Ar}$ ?

The successive ionization energies for an unknown element are $$\begin{aligned} I_{1} &=896 \mathrm{kJ} / \mathrm{mol} \\ I_{2} &=1752 \mathrm{kJ} / \mathrm{mol} \\ I_{3} &=14,807 \mathrm{kJ} / \mathrm{mol} \\\ I_{4} &=17,948 \mathrm{kJ} / \mathrm{mol} \end{aligned}$$ To which family in the periodic table does the unknown element most likely belong?

Identify the following three elements. a. The ground-state electron configuration is $[\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}$ b. The ground-state electron configuration is $[\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2}$ c. An excited state of this element has the electron configuration 1$s^{2} 2 s^{2} 2 p^{4} 3 s^{1}$

The electron affinities of the elements from aluminum to chlorine are \(-44,-120,-74,-200.4,\) and \(-384.7 \mathrm{kJ} / \mathrm{mol}\) respectively. Rationalize the trend in these values.

A carbon-oxygen double bond in a certain organic molecule absorbs radiation that has a frequency of \(6.0 \times 10^{13} \mathrm{s}^{-1}\) . a. What is the wavelength of this radiation? b. To what region of the spectrum does this radiation belong? c. What is the energy of this radiation per photon? per mole of photons? d. A carbon-oxygen bond in a different molecule absorbs radiation with frequency equal to \(5.4 \times 10^{13} \mathrm{s}^{-1} .\) Is this radiation more or less energetic?
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