Respuesta :
The de Broglie wavelength is 0.665 * [tex]10^{-14}[/tex] m.
Alpha particles are radioactive particles having two protons and two neutrons. It is the helium nucleus.
Δx * Δp = h / 2Π
where Δx = uncertainty in position
Δp = uncertainty in momentum
h = plank's constant
Δv = 0.1X10⁷ mi / h = 4.47 * [tex]10^{5}[/tex] m / s
we know that,
p = m * v
Therefore, Δp = m * Δv
= 6.6X10⁻²⁴ * 4.47 * [tex]10^{5}[/tex]
= 29.502 * [tex]10^{-19}[/tex] g m / s
= 29.502 * [tex]10^{-22}[/tex] kg m / s
p = m * v
= 6.6X10⁻²⁴ * 1.51 * [tex]10^{7}[/tex]
= 9.966 * [tex]10^{-20}[/tex] kg m / s
De Broglie wavelength:
λ = h / p
= 6.63 * [tex]10^{-34}[/tex] / 9.966 * [tex]10^{-20}[/tex]
λ= 0.665 * [tex]10^{-14}[/tex] m
Hence, the de Broglie wavelength is 0.665 * [tex]10^{-14}[/tex] m.
For more information on de Broglie's hypothesis click on the link below:
https://brainly.com/question/4851688
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