Respuesta :
Answer:
[tex]1.604\times 10^{-25} nm[/tex] is the de Broglie wavelength associated with a baseball that is moving with a velocity of 42 mph.
Explanation:
De-Broglie's wavelength, which is:
[tex]\lambda=\frac{h}{mv}[/tex]
where,
[tex]\lambda[/tex] = De-Broglie's wavelength = ?
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
m = mass of particle =
v = velocity of the particle
We have :
Mass of baseball = m = 220 g = 0.220 kg ( 1g = 0.001 kg)
Velocity of the base ball = v = 42mph =[tex]\frac{42}{2.237} m/s=18.78 m/s[/tex]
1 m/s = 2.237 mph
De-Broglie wavelength of the baseball at v:[tex]\lambda [/tex]
[tex]\lambda =\frac{6.626\times 10^{-34}Js}{0.220 kg\times 18.78 m/s}
[/tex]=1.604\times 10^{-34} m=1.604\times 10^{-25} nm[/tex]
[tex] 1 m = 10^9 nm[/tex]
[tex]1.604\times 10^{-25} nm[/tex] is the de Broglie wavelength associated with a baseball that is moving with a velocity of 42 mph.
