The radius of germanium $$\left( {Ge} \right)$$ nuclide is measured to be twice the radius of $$_4^9Be.$$ The number of nucleons in $$Ge$$ are
A.
74
B.
75
C.
72
D.
73
Answer :
72
Solution :
We use the formula,
$$R = {R_0}{A^{\frac{1}{3}}}$$
This represents relation between atomic mass and radius of the nucleus.
For berillium, $${R_1} = {R_0}{\left( 9 \right)^{\frac{1}{3}}}$$
For germanium, $${R_2} = {R_0}{A^{\frac{1}{3}}}$$
$$\eqalign{
& \frac{{{R_1}}}{{{R_2}}} = \frac{{{{\left( 9 \right)}^{\frac{1}{3}}}}}{{{{\left( A \right)}^{\frac{1}{3}}}}} \cr
& \Rightarrow \frac{1}{2} = \frac{{{{\left( 9 \right)}^{\frac{1}{3}}}}}{{{{\left( A \right)}^{\frac{1}{3}}}}} \cr
& \Rightarrow \frac{1}{8} = \frac{9}{A} \cr
& \Rightarrow A = 8 \times 9 = 72. \cr} $$
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is