Calculating IR Energy: 1.614 x 107 MHz

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SUMMARY

The energy of infrared radiation with a frequency of 1.614 x 107 MHz is calculated to be 6.44 kJ/mol. The calculation utilizes the equation E = nhv, where h is Planck's constant (6.63 x 10-34 J·s) and v is the frequency converted to Hz (1.614 x 1013 Hz). The conversion from joules to kilojoules is performed by dividing by 1000, and the final energy per mole is derived by multiplying the energy per photon by Avogadro's number (6.02 x 1023 atoms/mol). The correct understanding of unit conversions is essential for accurate calculations.

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  • Understanding of Planck's equation (E = nhv)
  • Knowledge of frequency conversion from MHz to Hz
  • Familiarity with unit conversions between joules and kilojoules
  • Basic grasp of Avogadro's number and its application in mole calculations
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  • Study the implications of Planck's constant in quantum mechanics
  • Learn about frequency and wavelength relationships in electromagnetic radiation
  • Explore detailed unit conversion techniques in physical chemistry
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Homework Statement



What is the energy (in kJ/mol) of infrared radiation that has a frequency of 1.614 x 107 MHz?

Homework Equations



E = nhv

The Attempt at a Solution



3 x 6.63 x 10^-34 (J x s) multiplied by 1.614 x 10^7 (MHz) multiplied by (10^6 Hz/1 MHz) multiplied by (1 kJ/ 1000J) = 3.210246 x 10^-20, which is wrong.
 
Last edited:
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E = hv
v=1.614x10^13 hz
h = 6.63x10^-34
E = 1.07x10^-20 J
E/1000 = kJ
E = 1.07x10^-23 kJ
per mole
1.07x10^-23 * (6.02 x 10^23 atoms/mol)
E = 6.44 kJ/mol

hope that is clear, any questions let me know :)
 


nickdk said:
E = hv
v=1.614x10^13 hz
h = 6.63x10^-34
E = 1.07x10^-20 J
E/1000 = kJ
E = 1.07x10^-23 kJ
per mole
1.07x10^-23 * (6.02 x 10^23 atoms/mol)
E = 6.44 kJ/mol

hope that is clear, any questions let me know :)

Yes that's the right answer! But I'm still having trouble understanding why, in terms of units, you can multiply the last step by (6.02 x 10^23 atoms/mol) and still get units of (kJ/mol). When did the unit of "atoms" appear?

Reflection: I made the mistake of not properly converting the Si-Units of "Mega-x" to "x" and neglected the aspect of moles.
 

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