a) Find the mass density of a proton, modeling it as a solid sphere of radius PS

glamrockqueen7

glamrockqueen7

Answered question

2021-09-14

a) Find the mass density of a proton, modeling it as a solid sphere of radius 1.00×1015m.
b) What If ? Consider a classical model of an electron as a uniform solid sphere with the same density as the proton. Find its radius.
c) Imagine that this electron possesses spin angular momentum lw=h2 because of classical rotation about the z axis. Determine the speed of a point on the equator of the electron.
d) State how this speed compares with the speed of light.

Answer & Explanation

Mitchel Aguirre

Mitchel Aguirre

Skilled2021-09-15Added 94 answers

Step 1
a) Let's us write the mass in terms of the density ρ and volume V:
m=ρV
The volume of a sphere is given by
V=43r3π
Therefore, we have
m=ρ43r3πρ=3m4r3π
Using the mass of the proton and the given radius we obtain
ρ=3(1.6727kg)4(1.001015m)3π
ρ=3.991017kgm3
b) We can express the radius from equation. (1):
r=(3m4ρπ)13
Inserting the electron mass and the density found in part (a) we get
r=(3(9.111031kg)4(3.991017kgm3)π13 
r=8.171017m
Step 3
c) The angular speed can be written as ω=vr so that we have
Lz=Iω=Ivr
The problem states that this is equal to h2Ivr=h2
v=rh2I
We found the radius in part (b) but what about the moment of inertia?
Observe that we have modeled an electron as a solid sphere which means that its moment of inertia can be expressed as:
I=25mr2
Thus, speed can be expressed as
v=rh225mr2=5h4mr
Inserting the numbers we get
v=5(1.0551034Js)4(9.111031kg)(8.171017m)
v=1.771017ms
Step 4
Answer: The speed found above is greater than the speed of light. This alone indicates that the model is incorrect.

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