What Limits Observation in Quantum Mechanics?

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SUMMARY

The discussion focuses on the limitations of observation in quantum mechanics, specifically addressing two problems related to an object of size 0.5 Angstrom. For the first problem, the longest-wavelength photon that can be used for observation is determined to be 0.5 Angstrom, while the second problem requires calculating the smallest-energy electron, which is found to be 602 eV. Key equations utilized include the Heisenberg uncertainty principle and the energy-frequency relationship E=hf. Participants emphasize the importance of understanding the relationship between a photon's momentum and its wavelength for accurate calculations.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the Heisenberg uncertainty principle.
  • Familiarity with the energy-frequency relationship, specifically E=hf.
  • Knowledge of photon momentum and its relationship to wavelength.
  • Basic concepts of electron energy calculations in quantum mechanics.
NEXT STEPS
  • Research the de Broglie wavelength and its implications in quantum mechanics.
  • Learn about the relationship between momentum and energy for particles, particularly electrons.
  • Study advanced applications of the Heisenberg uncertainty principle in experimental physics.
  • Explore the concept of wave-particle duality and its relevance in observation limits.
USEFUL FOR

Students of quantum mechanics, physics educators, and researchers interested in the foundational principles of observation limits in quantum systems.

blade_090
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1. Homework Statement [/b]
1)for an object of size 0.5 Angstrom, what is the longest-wavelength photon with which it can be observed?
2)for the object of problem 1, what is the smallest-energy electron which can be used to make the measurement?

Homework Equations


1)[tex]\Delta[/tex]p x [tex]\Delta[/tex] x [tex]\geq[/tex] [tex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s[/tex]/4(3.142)

2)[tex]\Delta[/tex]E x [tex]\Delta[/tex] t [tex]\geq[/tex] [tex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s[/tex]/4(3.142)

for problem 1,
0.5Angstrom is the wavelength.
i use the equation of E=hf to find the energy.
thn i use 2nd equation above to find [tex]\Delta[/tex] t
i use the [tex]\Delta[/tex] t [tex]\geq[/tex] [(lamda)^2]/4(pi)(speed of light)(delta lamda)
i did find the answer...bt somehow i feel like i did wrong

for problem 2,
i duno where i should start...
i tinking of using equation 2 bt I am not sure...whether energy equation from above is correct

the answer for prob 1 : 0.5Angstrom
the answer for prob 2 : 602eV
 
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blade_090 said:
1. Homework Statement [/b]
1)for an object of size 0.5 Angstrom, what is the longest-wavelength photon with which it can be observed?
2)for the object of problem 1, what is the smallest-energy electron which can be used to make the measurement?

Homework Equations


1)[tex]\Delta[/tex]p x [tex]\Delta[/tex] x [tex]\geq[/tex] [tex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s[/tex]/4(3.142)

2)[tex]\Delta[/tex]E x [tex]\Delta[/tex] t [tex]\geq[/tex] [tex]h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s[/tex]/4(3.142)
for problem 1,
0.5Angstrom is the wavelength.
i use the equation of E=hf to find the energy.

No, 0.5A is the size of the object. You're supposed to find the wavelength of the photon. If you were trying to find the de Broglie wavelength of the object, you can't use E=hf because that only applies to photons.

Can you find the momentum of the photon? If so, what's the relationship between a photon's momentum and its wavelength?

for problem 2,
i duno where i should start...
i tinking of using equation 2 bt I am not sure...whether energy equation from above is correct

Once you find the momentum from problem 1, you can apply it to an electron. How do you find an electron's energy from its momentum?
 

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