Find wavelength of a quantum of electromagnetic radiation

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SUMMARY

The wavelength of a quantum of electromagnetic radiation with an energy of 0.877 keV is calculated using the formula λ = hc/E. By converting the energy from electron-volts to joules, the correct wavelength is determined to be approximately 14.14 nm. The calculation involves using Planck's constant (6.62607 × 10-34 J·s) and the speed of light (2.99792 × 108 m/s). Proper unit conversion and significant figures are crucial for accuracy in the final result.

PREREQUISITES
  • Understanding of Planck's constant (6.62607 × 10-34 J·s)
  • Knowledge of the speed of light (2.99792 × 108 m/s)
  • Familiarity with energy conversion between electron-volts (eV) and joules (J)
  • Basic grasp of the equations E=hf and v=fλ
NEXT STEPS
  • Learn about energy conversion factors between electron-volts and joules
  • Study the derivation of the wavelength formula λ = hc/E
  • Explore the significance of significant figures in scientific calculations
  • Investigate applications of electromagnetic radiation in quantum mechanics
USEFUL FOR

Students in physics, particularly those studying quantum mechanics, as well as educators and anyone interested in the calculations involving electromagnetic radiation and its properties.

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Homework Statement


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


Homework Equations


E=hf
v=fλ
... λ=v/(E/h)

The Attempt at a Solution



When i solved, i got 1.413728e-9 nm... I have checked my units. can some just help and point me in the right direction
 
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v is usually written as c when one speaks of the speed of light in vacuum. Also, the double fraction reduces to:
<br /> \frac{c}{\frac{E}{h}} = \frac{h \, c}{E}<br />
For this answer, you need to know the conversion factor between an electron-volt (eV) and a joule as energy units. Do you know it?
 
Dickfore said:
v is usually written as c when one speaks of the speed of light in vacuum. Also, the double fraction reduces to:
<br /> \frac{c}{\frac{E}{h}} = \frac{h \, c}{E}<br />
For this answer, you need to know the conversion factor between an electron-volt (eV) and a joule as energy units. Do you know it?


Yes i did convert it but i still got it wrong
 
how did you convert it, and what did you get?
 
I did it again and i got 1.414E-8 ... 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
 
I didn't get that. What did you get for the energy in joules?
 
1.405109518e-16 J
 
This is correct. Now:
<br /> \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}}<br />

The product and ratio of the mantissas, gives:
<br /> \frac{6.626 \times 2.998}{1.4051} = 14.14<br />
The exponents sum up to -34 + 8 - (-16) = -10. You may read off the units from the above fraction fairly easily.

What should the answer be in scientific form?
 
so in nm, it would be 1.414
 
  • #10
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
Thank You
 

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