Unit and dimension about De broglie wave

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SUMMARY

The discussion centers on the de Broglie wave formula, specifically the relationship between Planck's constant (h) and momentum (p) in determining wavelength (λ). The formula λ = h/p is analyzed, with a focus on the unit conversion from joule-seconds (J·s) to meters (m). The breakdown of units shows that h/p simplifies to meters, confirming that the wavelength is expressed in SI units. This clarification aids in understanding the dimensional analysis involved in quantum mechanics.

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KT KIM
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From the de Broglie wave formula we know,
Rhamda=h/p

In actual examples of course the answer would be 'something [meters]'

I am having hardtime to understand how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit [m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.
 
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KT KIM said:
From the de Broglie wave formula we know,
Rhamda=h/p

In actual examples of course the answer would be 'something [meters]'

I am having hardtime to understand how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit [m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.
What is 1 Joule, expressed in SI base units?
 
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KT KIM said:
I am having hardtime how unit of h/mv
[J*s]/[kg]*[m/s] turn into wavelength unit [m]

I studied the Mass-Energy relation part earlier,
But still can't get the precise way how actually it works and modifes its unit.

Planck's constant is h -its units are joule-sec;
Joule is expressed as energy - its units are kg.m^2.s ^-2 ; so j--s will be kg.m^2.s^-1
so h/ p will be kg.m^2.s^-1 /(kg.m.s^-1 ) = m (in meters) which is the unit of Lambda the wavelength.
 
drvrm said:
Planck's constant is h -its units are joule-sec;
Joule is expressed as energy - its units are kg.m^2.s ^-2 ; so j--s will be kg.m^2.s^-1
so h/ p will be kg.m^2.s^-1 /(kg.m.s^-1 ) = m (in meters) which is the unit of Lambda the wavelength.
Thanks!
 

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