Question
The nuclear radius of $$_8{O^{16}}$$ is $$3 \times {10^{ - 15}}m.$$ If an atomic mass unit is $$1.67 \times {10^{ - 27}}kg,$$ then the nuclear density is approximately
A.
$$2.35 \times {10^{17}}g\,c{m^{ - 3}}$$
B.
$$2.35 \times {10^{17}}kg\,{m^{ - 3}}$$
C.
$$2.35 \times {10^{17}}g{m^{ - 3}}$$
D.
$$2.35 \times {10^{17}}kg\,m{m^{ - 3}}$$
Answer :
$$2.35 \times {10^{17}}kg\,{m^{ - 3}}$$
Solution :
For nucleus of $$_8{O^{16}}$$
Mass $$ = \left( {16} \right)\left( {1.67 \times {{10}^{ - 27}}} \right)kg$$
Volume $$ = \frac{4}{3}\pi {R^3}$$
$$ = \frac{4}{3}\pi {\left( {3 \times {{10}^{ - 15}}} \right)^3}{m^3} = 36\pi \times {10^{ - 45}}{m^3}$$
Density $$ = \frac{{{\text{mass}}}}{{{\text{volume}}}} = \frac{{16 \times 1.67 \times {{10}^{ - 27}}kg}}{{36\pi \times {{10}^{ - 45}}{m^3}}}$$
$$ = 2.35 \times {10^{17}}kg\,{m^{ - 3}}$$