# Question 88966

Dec 6, 2016

$1.6 \cdot {10}^{- 27} \text{kg}$

#### Explanation:

The idea here is that your particle has an associated de Broglie wavelength that depends on its momentum, $p$, which in turn depends on two things

• the mass of the particle
• the velocity of the particle

SIDE NOTE The problem wants you to use speed instead of velocity, but keep in mind that velocity and speed are not the same thing

More specifically, the momentum of the particle is given by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{p = m \cdot v}}}$

Here

• $m$ is the mass of the particle
• $v$ is its velocity

The de Broglie wavelength is given by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = \frac{h}{p}}}} \to$ the de Broglie wavelength

Here

• $p$ is the momentum of the particle
• $l a m \mathrm{da}$ is its wavelength
• $h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$

Now, the problem tells you that the aprticle is moving at 90.% the speed of light, which us usually taken to be

$c = 3.0 \cdot {10}^{8} {\text{m s}}^{- 1}$

This means that the speed of the particle will be equal to

$v = \frac{9.0}{10} \cdot c = \frac{9.0}{10} \cdot 3.0 \cdot {10}^{8} {\text{m s}}^{- 1}$

$v = 2.70 \cdot {10}^{8} {\text{m s}}^{- 1}$

Use the de Broglie wavelength to calculate the momentum of the particle

$l a m \mathrm{da} = \frac{h}{p} \implies p = \frac{h}{l a m \mathrm{da}}$

In order to be able to use the wavelength given to you, you must express Planck's constant using the fact that

${\text{1 J" = "1 kg m"^2 "s}}^{- 2}$

This means that you have

h = 6.626 * 10^(-34) "kg m s"^color(red)(cancel(color(black)(-2))) * color(red)(cancel(color(black)("s")))

$h = 6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

p = (6.626 * 10^(-34)"kg m"^color(red)(cancel(color(black)(2)))"s"^(-1))/(1.5 * 10^(-15)color(red)(cancel(color(black)("m")))) = 4.417 * 10^(-19)"kg m s"^(-1)

Now rearrange the equation that describes the particle's momentum to solve for $m$

$p = m \cdot v \implies m = \frac{p}{v}$

Plug in your values to find

m = (4.417 * 10^(-19) "kg" color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(2.70 * 10^(8)color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1))))) = color(darkgreen)(ul(color(black)(1.6 * 10^(-27)"kg")))#

The answer is rounded to two sig figs.

Dec 6, 2016

$1.643 \cdot {10}^{-} 27$kg.

#### Explanation:

I suppose we could use:
$m = \frac{h}{v \lambda}$

where:
$h$=planck's constant or $6.626 \cdot {10}^{-} 34 J \cdot s$
$m$=mass in kilogram
$v$=velocity in $m \cdot {s}^{-} 1$
$\lambda$=wavelength in meter

(If your going to use this formula, the units must be exact and if they are not you must convert them)

and for us,
$v = \left(2.998 \cdot {10}^{8}\right) \cdot 0.9 = 2.689 \cdot {10}^{8} m \cdot {s}^{-} 1$
$\lambda = 1.15 \cdot {10}^{-} 15 m$

$m = \frac{6.626 \cdot {10}^{-} 34 J \cdot s}{2.689 \cdot {10}^{8} m \cdot {s}^{-} 1 \cdot 1.5 \cdot {10}^{-} 15 m}$

The mass is of the particle is $1.643 \cdot {10}^{-} 27$kg.