## Unit conversion for DeBroglie?

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

Antonio Melgoza 2K
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### Unit conversion for DeBroglie?

This is from Example 1.6 in the textbook.
"Calculate the wavelength of a particle of mass 1 g traveling at 1 ms"
The book changes 1 gram into 1 x 10^(-3), but I don't understand how they got 1 x 10^(-3). Can someone explain please?

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### Re: Unit conversion for DeBroglie?

The question requires you to use the de Broglie relation: $\lambda = \frac{h}{p}$

Remember that units of h are: $J \cdot s = kg \cdot m^{2} \cdot s^{-1}$

Since Planck's constant is given to you in units of kg, you must work in kilograms which is $10^{3} g$.