Question #a20e7

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Question #a20e7

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Answer 1 · 2017-07-21T17:43:32

$2.70 \cdot 10^{- 12} m$ $2.70*10^(-12)m$

Explanation:

de Brogile wavelength - $λ = \frac{h}{p} = \frac{h}{v \cdot m_{e}}$ $lamda=h/p=h/(v*m_e)$ , where:
$λ$ $lamda$ = de brogile wavelength ( $m$ $m$ )
$h$ $h$ = Planck's constant ( $6 m 63 \cdot 10^{- 34} J$ $6m63*10^(-34)J$ $s$ $s$ )
$p$ $p$ = momentum of the particle ( $k g$ $kg$ $m$ $m$ $s^{- 1}$ $s^(-1)$ )
$v$ $v$ = velocity of the particle ( $m$ $m$ $s^{- 1}$ $s^(-1)$ )
$m_{e}$ $m_e$ = electron mass = ( $9.11 \cdot 10^{- 31} k g$ $9.11*10^(-31) kg$ )

$c \approx 3 \cdot 10^{8} m$ $c~~3*10^8m$ $s^{- 1}$ $s^(-1)$

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

$λ = \frac{6.63 \cdot 10^{- 34}}{(2.7 \cdot 10^{8}) (9.11 \cdot 10^{- 31})} \approx 2.70 \cdot 10^{- 12} m$ $lamda=(6.63*10^(-34))/((2.7*10^8)(9.11*10^(-31)))~~2.70*10^(-12)m$

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