Respuesta :
Answer:
For a: The De-Broglie wavelength of the given particle is [tex]2\times 10^{-10}m[/tex]
For b: The uncertainty in the position is [tex]3.56\times 10^{-14}m[/tex]
Explanation:
- For a:
To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:
[tex]\lambda=\frac{h}{mv}[/tex]
where,
[tex]\lambda[/tex] = De-Broglie's wavelength = ?
h = Planck's constant =
m = mass of the alpha particle =
v = velocity of alpha particle = [tex]3.4\times 10^7mi/h=1.52\times 10^7m/s[/tex] (Conversion factors used: 1 mile = 1609.34 m & 1 hr = 3600 s)
Putting values in above equation, we get:
[tex]\lambda=\frac{6.6\times 10^{-34}Js}{6.6\times 10^{-27}kg\times 1.52\times 10^7m/s}\\\\\lambda=6.58\times 10^{-15}m[/tex]
Hence, the De-Broglie wavelength of the given particle is [tex]2\times 10^{-10}m[/tex]
- For b:
The equation representing Heisenberg's uncertainty principle follows:
[tex]\Delta x.\Delta p\geq \frac{h}{2\pi}[/tex]
where,
[tex]\Delta x[/tex] = uncertainty in position = ?
[tex]\Delta p[/tex] = uncertainty in momentum = [tex]m\Delta v[/tex]
m = mass of the alpha particle = [tex]6.6\times 10^{-27}kg[/tex]
[tex]\Delta v[/tex] = uncertainty in speed = [tex]0.1\times 10^7mi/hr=4.47\times 10^5m/s[/tex]
h = Planck's constant = [tex]6.6\times 10^{-34}kgm^2/s[/tex]
Putting values in above equation, we get:
[tex]\Delta x=\frac{6.6\times 10^{-34}kgm^2/s}{2\times 3.14 \times 6.6\times 10^{-27}\times kg 4.47\times 10^{5}m/s}\\\\\Delta x=3.56\times 10^{-14}m[/tex]
Hence, the uncertainty in the position is [tex]3.56\times 10^{-14}m[/tex]
