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Question
From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).
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Solution
We have the expression for nuclear radius as:
R = R0A1/3
Where,
R0 = Constant.
A = Mass number of the nucleus
Nuclear matter density, `rho = "Mass of the nucleus"/"Volume of the nucleus"`
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
`therefore rho = "mA"/(4/3 pi"R"^3) = "3mA"/(4pi ("R"_0 "A"^(1/3))^3) = (3"mA")/(4pi"R"_0^3 "A") = "3m"/(4pi"R"_0^3)`
Hence, the nuclear matter density is independent of A. It is nearly constant.
RELATED QUESTIONS
Answer the following question.
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The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb − mC − md]c² where the masses refer to the respective nuclei. Calculate the Q-value of the reaction and state whether this reaction is exothermic or endothermic.
\[_1^1\mathrm{H}+_1^3\mathrm{H}\rightarrow_1^2\mathrm{H}+_1^2\mathrm{H}\]
Atomic masses are given to be
\[\mathrm{m(}_{1}^{1}\mathrm{H)}=1.007825\mathrm{u}\]
\[\mathrm{m(}_{1}^{2}\mathrm{H)}=2.014102\mathrm{u}\]
\[\mathrm{m(}_{1}^{3}\mathrm{H})=3.016049\mathrm{u}\]
Assertion (A): All nuclei are not of the same size.
Reason (R): The size of the nucleus depends on atomic mass.
A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to 2:1. What will be the ratio of their nuclear size (nuclear radius)?
The value of the radius constant \[R_0\] used in the nuclear radius formula is approximately:
The minimum distance reached by an alpha particle approaching a positively charged nucleus before being turned back is called the:
If the mass number \[A\] of a nucleus increases, its radius increases as \[A^{1/3}\]. This implies that:
Since the nucleus is approximately spherical with \[R = R_0 A^{1/3}\], its volume \[V = \frac{4}{3}\pi R^3\] is proportional to:
The approximate density of nuclear matter is:
