The mass density of a nucleus varies with mass number $$A$$ as
A.
$${A^2}$$
B.
$$A$$
C.
constant
D.
$$\frac{1}{A}$$
Answer :
constant
Solution :
Density of nuclear matter is the ratio of mass of nucleus and its volume.
If $$m$$ is average mass of a nucleon and $$R$$ is the nuclear radius, then mass of nucleus $$= mA,$$ where $$A$$ is the mass number of the element.
Volume of nucleus $$ = \frac{4}{3}\pi {R^3}$$
$$\eqalign{
& = \frac{4}{3}\pi {\left( {{R_0}{A^{\frac{1}{3}}}} \right)^3} \cr
& = \frac{4}{3}\pi R_0^3A \cr} $$
As density of nuclear matter $$ = \frac{{{\text{mass of nucleus}}}}{{{\text{volume of nucleus}}}}$$
$$\eqalign{
& \rho = \frac{{mA}}{{\frac{4}{3}\pi R_0^3A}} \cr
& \therefore \rho = \frac{{3m}}{{4\pi R_0^3}} \cr} $$
As $$m$$ and $${{R_0}}$$ are constants, therefore density $$\rho $$ of nuclear matter is constant.
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