De Broglie Equation Dividing J*s by kg*m/s

$\lambda=\frac{h}{p}$

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marcus lin 1E
Posts: 54
Joined: Fri Sep 28, 2018 12:28 am

De Broglie Equation Dividing J*s by kg*m/s

So for De Broglie's Equation, you have to divide planck's constant (h) by momentum. But the units of h are in J*s and momentum is in kg*m/s, can someone explain to me how the units cancel out to get a unit appropriate for wavelength?

Harshita Talkad 4L
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: De Broglie Equation Dividing J*s by kg*m/s

During Friday's lecture, Professor Lavelle said that 1 J= 1 kg.m^2.s^(-2). In the equation, the units are (J.s)/(kg.m/s), so if you plug in the units for J, all the units except for m will cancel out, which gives wavelength in meters.

shaunajava2e
Posts: 66
Joined: Fri Sep 28, 2018 12:26 am

Re: De Broglie Equation Dividing J*s by kg*m/s

1 J= 1 kg.m^2.s^(-2)
the units in the de broglie equation are: (J.s)/(kg.m/s)
this becomes: [(1 kg.m^2.s^(-2))(s)]/(kg.m/s)
the kgs on both sides cancel out
there is a s^(-2) and s on top, and a s on the bottom, these cancel out as well
the m^2 on top cancels out with one m on the bottom to end with a single m which is your resulting units

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