Question

The radius of Ge nuclide is measured to be twice the radius of $ {}_4B{e^9} $ . The number of nucleon in Ge are
(A) 72
(B) 73
(C) 74
(D) 75

Answer 1

Hint: The nucleon number of $ {}_4B{e^9} $ is 9. The radius of a nuclide is directly proportional to the third root of its mass number or nucleon number. So using this formula we can find the number of nucleons in Ge nuclei.

Formula used: In this solution we will be using the following formula;
$= k{A^{\dfrac{1}{3}}} $ where $ R $ is the radius of the nuclide of an atom, $ A $ is the nucleon number of the atom, and $ k $ is a constant of proportionality.

Complete step by step solution:
The radius of the nucleus of an atom is determined by the number of nucleons in the nucleus. The higher the number, the larger the nucleus. Specifically, the radius is directly proportional to the cube root of the mass number (which is the same as the nucleon number). Hence
$ R \propto {A^{\dfrac{1}{3}}} \Rightarrow R = k{A^{\dfrac{1}{3}}} $ where $ R $ is the radius of the nuclide of an atom, $ A $ is the nucleon number of the atom, and $ k $ is a constant of proportionality.
From the equation above, we have that
 $\Rightarrow \dfrac{R}{{{A^{\dfrac{1}{3}}}}} = k $
Since this constant is same for all atoms, so we can write,
 $\Rightarrow \dfrac{{{R_{Ge}}}}{{A_{Ge}^{\dfrac{1}{3}}}} = \dfrac{{{R_{Be}}}}{{A_{Be}^{\dfrac{1}{3}}}} $
Rearranging, we have that
 $\Rightarrow \dfrac{{{R_{Ge}}}}{{{R_{Be}}}} = \dfrac{{A_{Ge}^{\dfrac{1}{3}}}}{{A_{Be}^{\dfrac{1}{3}}}} $
 $\Rightarrow \dfrac{{{R_{Ge}}}}{{{R_{Be}}}} = {\left( {\dfrac{{{A_{Ge}}}}{{{A_{Be}}}}} \right)^{\dfrac{1}{3}}} $
Cubing both sides, we have that
 $\Rightarrow \dfrac{{R_{Ge}^3}}{{R_{Be}^3}} = \dfrac{{{A_{Ge}}}}{{{A_{Be}}}} $
Making $ {A_{Ge}} $ subject of the formula, we have
 $\Rightarrow {A_{Ge}} = {\left( {\dfrac{{{R_{Ge}}}}{{{R_{Be}}}}} \right)^3}{A_{Be}} $
According to the question the radius of germanium is twice that of the radius of the Beryllium thus, $ {R_{Ge}} = 2{R_{Be}} $
Hence, we get
 $\Rightarrow {A_{Ge}} = {\left( {\dfrac{{2{R_{Be}}}}{{{R_{Be}}}}} \right)^3}{A_{Be}} $
On cancelling $ {R_{Be}} $ and substituting the value of the mass number of $ {}_4B{e^9} $
 $\Rightarrow {A_{Ge}} = {\left( 2 \right)^3} \times 9 $
Thus, $ {A_{Ge}} = 8 \times 9 = 72nucleons $
Hence, the correct answer is option A.

Note:
Alternatively, we can calculate for the expression of $ k $ from $ \dfrac{{{R_{Be}}}}{{{A^{\dfrac{1}{3}}}}} = k $
$\Rightarrow \dfrac{{{R_{Be}}}}{{{A^{\dfrac{1}{3}}}}} = k $
$\Rightarrow k = \dfrac{{{R_{Be}}}}{{{9^{\dfrac{1}{3}}}}} $
Then insert the expression into $ {R_{Ge}} = kA_{Ge}^{\dfrac{1}{3}} $ .
We have that
 $\Rightarrow {R_{Ge}} = \dfrac{{{R_{Be}}}}{{{9^{\dfrac{1}{3}}}}} \times A_{Ge}^{\dfrac{1}{3}} \Rightarrow 2{R_{Be}} = \dfrac{{{R_{Be}}}}{{{9^{\dfrac{1}{3}}}}} \times A_{Ge}^{\dfrac{1}{3}} $
Hence, by cancelling out the common term $ {R_{Be}} $ , and multiplying both sides by $ {9^{\dfrac{1}{3}}} $ , we have that
 $\Rightarrow 2 \times {9^{\dfrac{1}{3}}} = A_{Ge}^{\dfrac{1}{3}} $
Again, by cubing both sides, we have that
 $\Rightarrow {A_{Ge}} = {\left( {2 \times {9^{\dfrac{1}{3}}}} \right)^3} = {2^3} \times 9 $
Hence, $ {A_{Ge}} = 72 $ nucleons. The result is identical to the result in the step by step solution.