Calculating the De Broglie Wavelength of an Accelerated Particle

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Homework Help Overview

The discussion revolves around calculating the de Broglie wavelength of a particle with mass m and charge q that is accelerated across a potential difference V, specifically in a non-relativistic context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the de Broglie wavelength using the relationship between kinetic energy and potential energy, leading to a proposed formula. Some participants engage in unit analysis to verify the dimensional consistency of the derived expression.

Discussion Status

The discussion includes attempts to validate the proposed formula through dimensional analysis. While some participants express confidence in the findings, there is no explicit consensus on the correctness of the solution, and the conversation remains open-ended.

Contextual Notes

Participants are operating under the assumption of non-relativistic conditions and are exploring the implications of the potential difference on the particle's energy and wavelength.

Safinaz
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Homework Statement

A particle of mass m and charge q is accelerated across a potential dierence V to a non-relativistic velocity. What is the de Broglie wavelength of this particle?

Homework Equations



Is it

upload_2015-9-20_0-12-18.png


The Attempt at a Solution



I think it's
## \frac{h}{\sqrt{2mqV}} ##, because ## qV = \frac{p^2}{2m}## , or : P.E (at rest) = K.E (after acceleration), so that ## p= \sqrt{2mqV}, \lambda= \frac{h}{p}= \frac{h}{\sqrt{2mqV}}##
 
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You can boost your confidence by using the fact that the provided options all have different units.
 
## (\frac{h}{\sqrt{2mqV}} )^2 : \frac{J^2 . s^2}{kg.C.volt} = \frac{kg^2 . m^4}{s^2} \times \frac{A . s^3}{kg. C. Kg. m^2} ##
## = m^2 \times \frac{C/s. s}{C} = m^2 ##
 
Well then you have found the answer.
 
Thanks :)
 

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