X rays have wavelengths on the order of 1 3 10210 m. calcu‑ late the energy of 1.0 3 10210 m x rays in units of kilojoules per mole of x rays. (1 mol x rays 5 6.022 3 1023 x rays.) am radio waves have wavelengths on the order of 1 3 104 m. cal‑ culate the energy of 1.0 3 104 m radio waves in units of kilo‑ joules per mole of radio waves. consider that the bond energy of a carbon–carbon single bond found in organic compounds is 347 kj/mol. would x rays and/or radio waves be able to disrupt organic compounds by breaking carbon–carbon single bonds?

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04-09-2019
Physics

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X rays have wavelengths on the order of 1 3 10210 m. calcu‑ late the energy of 1.0 3 10210 m x rays in units of kilojoules per mole of x rays. (1 mol x rays 5 6.022 3 1023 x rays.) am radio waves have wavelengths on the order of 1 3 104 m. cal‑ culate the energy of 1.0 3 104 m radio waves in units of kilo‑ joules per mole of radio waves. consider that the bond energy of a carbon–carbon single bond found in organic compounds is 347 kj/mol. would x rays and/or radio waves be able to disrupt organic compounds by breaking carbon–carbon single bonds?

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cryssatemp cryssatemp · Answer 1 · 2019-09-12T21:02:16+02:00

Answers:

a) [tex]E_{Xrays}=1.19(10)^{6} kJ/mol[/tex]

b) [tex]E_{AM}= 1.19(10)^{-8} kJ/mol[/tex]

c) X rays have the energy to disrupt organic compounds by breaking carbon–carbon single bonds, but AM radio waves not.

Explanation:

The energy of a photon in Joules is given by the following formula:

[tex]E=\frac{hc}{\lambda}[/tex] (1)

However, if we want to know this energy in units of [tex]J/mol[/tex], we have to use the following formula:

[tex]E=\frac{hc}{\lambda} NA[/tex] (2)

a) Now, in the case of X rays:

[tex]E_{Xrays}=\frac{hc}{\lambda_{Xrays}} NA[/tex] (3)

Where:

[tex]E_{Xrays}[/tex] is the energy of X rays in Joules ([tex]J[/tex])

[tex]h=6.6260701509 (10)^{-34} J.s[/tex] is the Planck constant

[tex]c=3(10)^{8}m/s[/tex] is the speed of light

[tex]\lambda_{Xrays}=1(10)^{-10}m[/tex] is the wavelength of X rays

[tex]NA=6.0221(10)^{23}/mol[/tex] is the Avogadro's number

[tex]E_{Xrays}=\frac{(6.6260701509 (10)^{-34} J.s)(3(10)^{8}m/s)}{1(10)^{-10}m}(6.0221(10)^{23}/mol)[/tex] (4)

[tex]E_{Xrays}=1.19(10)^{9}J/mol=1.19(10)^{6} kJ/mol[/tex] (5) This is the energy of X rays in units of kilojoules per mole

b) For AM radio waves:

[tex]E_{AM}=\frac{hc}{\lambda_{AM}} NA[/tex] (6)

Where:

[tex]E_{AM}[/tex] is the energy of AM radio waves in Joules ([tex]J[/tex])

[tex]\lambda_{AM}=1(10)^{4}m[/tex] is the wavelength of AM wavelength

[tex]E_{AM}=\frac{(6.6260701509 (10)^{-34} J.s)(3(10)^{8}m/s)}{1(10)^{4}m} 6.0221(10)^{23}/mol[/tex] (7)

[tex]E_{AM}=1.19(10)^{-5} J/mol= 1.19(10)^{-8} kJ/mol[/tex] (8) This is the energy of AM radio waves in units of kilojoules per mole

c) Which is able to disrupt carbon–carbon single bonds?

If we compare the energy of both (X rays and AM radio waves) with the energy of a carbon–carbon single bond ([tex]347 kJ/mol[/tex]), we will have the following:

[tex]1.19(10)^{6} kJ/mol > 1.19(10)^{-8} kJ/mol > 347 kJ/mol[/tex]

This means the energy of X rays is greater than the energy of AM radio waves and greater than the energy of carbon–carbon single bond as well:

[tex]E_{Xrays} > E_{AM} > E_{carbon}[/tex]

In addition, this means X rays have the energy to disrupt organic compounds by breaking carbon–carbon single bonds, but AM radio waves not.