Calculating the De Broglie Wavelength of an Accelerated Particle

AI Thread Summary
The de Broglie wavelength of a particle with mass m and charge q, accelerated through a potential difference V, is calculated using the formula λ = h/√(2mqV). This equation derives from equating the potential energy gained (qV) to the kinetic energy (KE = p²/2m). The momentum p is expressed as p = √(2mqV), leading to the wavelength formula. The discussion also confirms the dimensional analysis of the units, ensuring the result is consistent. The final answer is validated through both the formula and unit checks.
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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

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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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