Question #a20e7

Jul 21, 2017

$2.70 \cdot {10}^{- 12} m$

Explanation:

de Brogile wavelength - $l a m \mathrm{da} = \frac{h}{p} = \frac{h}{v \cdot {m}_{e}}$, where:
$l a m \mathrm{da}$ = de brogile wavelength ($m$)
$h$ = Planck's constant ($6 m 63 \cdot {10}^{- 34} J$ $s$)
$p$ = momentum of the particle ($k g$ $m$ ${s}^{- 1}$)
$v$ = velocity of the particle ($m$ ${s}^{- 1}$)
${m}_{e}$ = electron mass = ($9.11 \cdot {10}^{- 31} k g$)

$c \approx 3 \cdot {10}^{8} m$ ${s}^{- 1}$

$0.9 c = 2.7 \cdot {10}^{8} m$ ${s}^{- 1}$

$l a m \mathrm{da} = \frac{6.63 \cdot {10}^{- 34}}{\left(2.7 \cdot {10}^{8}\right) \left(9.11 \cdot {10}^{- 31}\right)} \approx 2.70 \cdot {10}^{- 12} m$