Question
The nuclear radius of a nucleus with nucleon number 16 is $$3 \times {10^{ - 15}}m.$$ Then, the nuclear radius of a nucleus with nucleon number 128 is
A.
$$3 \times {10^{ - 15}}m$$
B.
$$1.5 \times {10^{ - 15}}m$$
C.
$$6 \times {10^{ - 15}}m$$
D.
$$4.5 \times {10^{ - 15}}m$$
Answer :
$$6 \times {10^{ - 15}}m$$
Solution :
Radius $$R$$ as a nucleus changes with the nucleon number $$A$$ of the nucleus as
$$R = 1.3 \times {10^{ - 15}} \times {A^{\frac{1}{3}}}m$$
Hence, $$\frac{{{R_2}}}{{{R_1}}} = {\left( {\frac{{{A_2}}}{{{A_1}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{128}}{{16}}} \right)^{\frac{1}{3}}} = {\left( 8 \right)^{\frac{1}{3}}} = 2$$
$$\therefore {R_2} = 2{R_1} = 2\left( {3 \times {{10}^{ - 15}}} \right)m = 6 \times {10^{ - 15}}m$$