1) [tex]1.05\cdot 10^{-9} m[/tex]
The wavelength of the matter wave (also called de Broglie wavelength) of an object is given by
[tex]\lambda=\frac{h}{mv}[/tex]
where
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
m is the mass of the object
v is its velocity
For a proton, we have:
[tex]m=1.67\cdot 10^{-27} kg[/tex]
and the velocity of this proton is
[tex]v=377 m/s[/tex]
So, its de Broglie's wavelength is:
[tex]\lambda=\frac{6.63\cdot 10^{-34}}{(1.67\cdot 10^{-27})(377)}=1.05\cdot 10^{-9} m[/tex]
2) [tex]1.11\cdot 10^{-38} m[/tex]
We can use again the same equation:
[tex]\lambda=\frac{h}{mv}[/tex]
where in this case we have:
m = 159 kg is the mass of the astronaut + spacesuit
v = 377 m/s is the velocity of the astronaut
Substituting into the equation,
[tex]\lambda=\frac{6.63\cdot 10^{-34}}{(159)(377)}=1.11\cdot 10^{-38} m[/tex]
3) [tex]3.70\cdot 10^{-63} m[/tex]
Similarly, we can use the same equation:
[tex]\lambda=\frac{h}{mv}[/tex]
where in this case we have:[tex]m=5.98\cdot 10^{24}kg[/tex] is the Earth's mass
[tex]v=30 km/s = 30000 m/s[/tex] is the velocity of the Earth around the Sun
Substituting,
[tex]\lambda=\frac{6.63\cdot 10^{-34}}{(5.98\cdot 10^{24})(30000)}=3.70\cdot 10^{-63} m[/tex]
This question can be used using the concept of de Broglie's wavelength and matter-wave.
a) Wavelength of matter-wave associated is "1.052 x 10⁻⁶ mm".
b) Wavelength of matter-wave associated with an astronaut is "1.105 x 10⁻³⁵ mm".
c) Wavelength of matter-wave is associated with Earth "3.68 x 10⁻⁶³ m".
The concept of matter-wave states that every matter has a certain wavelength due to its motion. This wavelength can be found using the formula of de Broglie's wavelength.
[tex]\lambda = \frac{h}{mv}[/tex]
where,
λ = de Broglie's Wavelength
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
m = mass
v = speed
a)
For the proton:
m = 1.67 x 10⁻²⁷ kg
v = 377 m/s (correction, double written in question)
[tex]\lambda = \frac{6.625\ x\ 10^{-34}\ J.s}{(1.67\ x\ 10^{-27}\ kg)(377\ m/s)}[/tex]
λ = 1.052 x 10⁻⁹ m = 1.052 x 10⁻⁶ mm
b)
For the astronaut:
m = 159 kg (correction, double written in question)
v = 377 m/s (correction, double written in question)
[tex]\lambda = \frac{6.625\ x\ 10^{-34}\ J.s}{(159\ kg)(377\ m/s)}[/tex]
λ = 1.105 x 10⁻⁻³⁸ m = 1.105 x 10⁻³⁵ mm
c)
For the Earth moving around Sun:
m = 6 x 10²⁴ kg
v = 29951.68 m/s
[tex]\lambda = \frac{6.625\ x\ 10^{-34}\ J.s}{(6\ x\ 10^{24}\ kg)(29951.68\ m/s)}\\\\[/tex]
λ = 3.68 x 10⁻⁶³ m
Learn more about de Broglie's wavelength here:
https://brainly.com/question/17295250?referrer=searchResults
The attached picture shows the formula of de Broglie's wavelength.
