Answer:
The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}4.6656×10
−14
.
The density of a proton is 6.278\times 10^{14} g/cm^36.278×10
14
g/cm
3
.
Explanation:
Diameter of the atom ,d = 2.50 Å
Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å
Volume of the sphere= \frac{4}{3}\pi r^3
3
4
πr
3
Volume of atom = V
V=\frac{4}{3}\pi r^3V=
3
4
πr
3
..[1]
Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA9.00×10
−5
A
˚
Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA0.5×9.00×10
−5
A
˚
=4.5×10
−5
A
˚
Volume of nucleus = V'
V=\frac{4}{3}\pi r'^3V=
3
4
πr
′3
..[2]
Dividing [2] by [1]
\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}
V
V
′
=
3
4
πr
3
3
4
πr
′3
=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}=
r
3
r
′3
=
(1.25
A
˚
)
3
(4.5×10
−5
A
˚
)
3
\frac{V'}{V}=4.6656\times 10^{-14}
V
V
′
=4.6656×10
−14
The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}4.6656×10
−14
.
Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm1.72×10
−15
m=1.72×10
−13
cm
1 m = 100 cm
Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm0.5×1.72×10
−13
cm=8.6×10
−14
cm
Volume of the sphere= \frac{4}{3}\pi r^3
3
4
πr
3
Volume of atom = V
V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3V=
3
4
×3.14×(8.6×10
−14
cm)
3
=2.664×10
−39
cm
3
Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g1.0073×1.66054×10
−24
g
1 amu = 1.66054\times 10^{-24} g1amu=1.66054×10
−24
g
Density of the proton : d
d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3d=
V
m
=
2.664×10
−39
cm
3
1.0073×1.66054×10
−24
g
=6.278×10
14
g/cm
3
The density of a proton is 6.278\times 10^{14} g/cm^36.278×10
14
g/cm
3
.
Explanation:
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