Calculate the wavelength of electrons traveling at 1.15x10^5

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Homework Help Overview

The problem involves calculating the wavelength of electrons traveling at a specified speed, utilizing the de Broglie wavelength formula. The context is rooted in quantum mechanics and wave-particle duality.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the wavelength using the provided formula and constants but questions the arithmetic leading to a discrepancy with the textbook answer. Participants discuss the appropriateness of unit conversion to nanometers.

Discussion Status

Some participants have offered guidance on re-evaluating the arithmetic involved in the calculation. There is an acknowledgment of the typical use of nanometers for small wavelengths, and further exploration of related energy calculations has been introduced.

Contextual Notes

Participants are considering the implications of their calculations in relation to energy content comparisons, and there is an ongoing discussion about the accuracy of the original poster's arithmetic and the assumptions made in the problem setup.

Anna Davis
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Homework Statement


Calculate the wavelength of electrons traveling at 1.15x10^5 m/s

Homework Equations


λ = h/m*v

Given values:
h=6.626x10^-34 J*s
m=9.11x10^-31kg
v=1.15x10^5 m/s

The Attempt at a Solution


I converted J*s to SI units (kg⋅m2⋅s−2) and then substituted out all known values, to get:
λ = (6.626x10^-34 kg⋅m2⋅s−2)/(9.11x10^-31kg)(1.15x10^5 m/s)

After canceling out like units and doing the math, I get:
λ = 6.00x10^-9m

The text gives the solution as 6.33nm. I can't figure out how they got there (and why they converted to nm?).

Is this an error in the text or (more likely) in my logic/calcuations?

Any/all input is appreciated. Thanks!
 
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Anna Davis said:
The text gives the solution as 6.33nm. I can't figure out how they got there (and why they converted to nm?).
Your expression is correct, just redo the arithmetic. For wavelengths that small, nm is a typical unit.
 
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Ah, good to know about nm... And I'll do the math again (hopefully better!). Thanks :)
 
If interested, and using the wavelength determined in the above problem, calculate the amount of energy in 1 gram of electrons moving at the same speed. (See if you can get ∆E = 1.9 x 104Kj/mole e-'s = 3.45 x 107Kj/g e-'s. Compare: A candle flame generating q ~ 40.7 Kj/g from burning paraffin wax to the energy content of a gram of electrons traveling at 1.15 x 105m/s. => You might find it very surprising. You are definitely on the right track. Good job!
 

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