# What is the speed of a 50 g rock if its de Broglie wavelength is #3.32*10^-34# m?

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#h = 6.63*10^-34 J*s#

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that all matter can behave like *waves*, as given by the **de Broglie hypothesis**.

The relationship between the *wavelength* of a massive particle, which is known as the **de Broglie wavelength**, and the **momentum** of the particle is described by the following equation

#color(blue)(|bar(ul(color(white)(a/a) lamda * p = h color(white)(a/a)|)))#

Here

*de Broglie wavelength*

*momentum* of the particle

**Planck's constant**, given to you as

Your first goal here will be to use this equation to find the momentum of the particle. Rearrange to solve for

#lamda * p = h implies p = h/(lamda)#

Before plugging in your values, use the fact that you have

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 J" = 1 "kg m"^2"s"^(-2))color(white)(a/a)|)))#

to write the Planck constant as

#h = 6.63 * 10^(-34)"kg m"^2"s"^(-2) * "s"#

#h = 6.63 * 10^(-34)"kg m"^2"s"^(-1)#

Now plug in your values to find the momentum of the particle

#p = (6.63 * 10^(-34)"kg m"^color(red)(cancel(color(black)(2))) "s"^(-1))/(3.32 * 10^(-34)color(red)(cancel(color(black)("m")))) = "1.997 kg m s"^(-1)#

Now, the momentum of a particle is given by the product between its **mass**, **velocity**,

#color(blue)(|bar(ul(color(white)(a/a)p = m * v color(white)(a/a)|)))#

Convert the mass of the particle from *grams* to *kilograms*

#50 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.050 kg"#

Plug in your values to find the *velocity* of the particle

#p =m * v implies v = p/m#

#v = (1.997 color(red)(cancel(color(black)("kg")))"m s"^(-1))/(0.050color(red)(cancel(color(black)("kg")))) = "39.94 m s"^(-1)#

Now, you don't have any information about the **direction** of motion for your particle, which means that you can say that the **magnitude** of the velocity, which is essentially the *speed* of the particle, is equal to

#v = color(green)(|bar(ul(color(white)(a/a)color(black)("40 m s"^(-1))color(white)(a/a)|)))#

The answer must be rounded to one **sig fig**, the number of sig figs you have for the mass of the particle.