# Question ee4ab

Aug 17, 2017

$l a m {\mathrm{da}}_{\text{matter}} = 10 \cdot \sqrt{\frac{h}{m}}$

#### Explanation:

This problem is more or less an exercise in algebraic manipulation because all you have to do here is to use the equation that describes the de Broglie wavelength of the particle

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

Here

• $l a m {\mathrm{da}}_{\text{matter}}$ is its de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$
• $m$ is the mass of the particle
• $v$ is its velocity

In your case, you know that

$l a m {\mathrm{da}}_{\text{matter}} = 100 \cdot v$

This implies that

$v = l a m {\mathrm{da}}_{\text{matter}} / 100$

Plug this into the above equation to get

lamda_"matter" = h/(m * lamda_"matter"/100)#

At this point, all you have to do is to isolate $l a m {\mathrm{da}}_{\text{matter}}$ by using algebraic manipulation.

$l a m {\mathrm{da}}_{\text{matter" = 100 * h/(m * lamda_"matter}}$

$l a m {\mathrm{da}}_{\text{matter" * m * lamda_"matter}} = 100 \cdot h$

$l a m {\mathrm{da}}_{\text{matter}}^{2} = \frac{100 \cdot h}{m}$

Therefore, you can say that

$l a m {\mathrm{da}}_{\text{matter}} = \sqrt{\frac{100 \cdot h}{m}}$

which is equivalent to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{l a m {\mathrm{da}}_{\text{matter}} = 10 \cdot \sqrt{\frac{h}{m}}}}}$