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Find wavelength of a quantum of electromagnetic radiation 
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#1
Mar2612, 08:03 PM

P: 47

1. The problem statement, all variables and given/known data
A quantum of electromagnetic radiation has an energy of 0.877 keV. What is its wavelength? The speed of light is 2.99792 × 10 8 m/s, and Planck’s constant is 6.62607 × 10−34J · s. Answer in units of nm 2. Relevant equations E=hf v=fλ ... λ=v/(E/h) 3. The attempt at a solution When i solved, i got 1.413728e9 nm... I have checked my units. can some just help and point me in the right direction 


#2
Mar2612, 08:29 PM

P: 3,014

v is usually written as c when one speaks of the speed of light in vacuum. Also, the double fraction reduces to:
[tex] \frac{c}{\frac{E}{h}} = \frac{h \, c}{E} [/tex] For this answer, you need to know the conversion factor between an electronvolt (eV) and a joule as energy units. Do you know it? 


#3
Mar2612, 08:32 PM

P: 47

Yes i did convert it but i still got it wrong 


#4
Mar2612, 08:33 PM

P: 3,014

Find wavelength of a quantum of electromagnetic radiation
how did you convert it, and what did you get?



#5
Mar2612, 08:36 PM

P: 47

I did it again and i got 1.414E8 ... and i think that is in meters. Am i right??
so that means that the answer is14.14nm _______________________________________________________________________ _ I used plancks constant in eV's. Its on the ap equation sheet 


#6
Mar2612, 08:41 PM

P: 3,014

I didn't get that. What did you get for the energy in joules?



#7
Mar2612, 08:45 PM

P: 47

1.405109518e16 J



#8
Mar2612, 08:51 PM

P: 3,014

This is correct. Now:
[tex] \frac{h \, c}{E} = \frac{6.626 \times 10^{34} \, \mathrm{J} \cdot \mathrm{s} \times 2.998 \times 10^8 \, \mathrm{m} \cdot \mathrm{s}^{1}}{1.4051 \times 10^{16} \, \mathrm{J}} [/tex] The product and ratio of the mantissas, gives: [tex] \frac{6.626 \times 2.998}{1.4051} = 14.14 [/tex] The exponents sum up to [itex]34 + 8  (16) = 10[/itex]. You may read off the units from the above fraction fairly easily. What should the answer be in scientific form? 


#9
Mar2612, 08:58 PM

P: 47

so in nm, it would be 1.414



#10
Mar2612, 09:03 PM

P: 3,014

yes, except that you need to use as many significant figures, as there are in variable with the least number of significant figures given in the problem. Fundamental constants are usually known to a lot of significant figures.



#11
Mar2612, 09:21 PM

P: 47

Thank You



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