Question 141:
easyThe mass number of He is 4 and that of sulphur is 32. The radius of sulphur nucleus is larger than that of helium by the factor of
(1995)
The radius of a nucleus is proportional to the cube root of its mass number ($R \propto A^{1/3}$). The ratio $R_S / R_{He} = (32 / 4)^{1/3} = (8)^{1/3} = 2$. Therefore, the radius is larger by a factor of 2.
Question 142:
easyThe mass density of a nucleus varies with mass number A as
(1992)
Nuclear density is defined as mass per unit volume. Since mass is proportional to A and volume is proportional to $R^3 \propto A$, the density is proportional to $A/A = 1$. It is a constant independent of A.
Question 143:
easyThe constituents of atomic nuclei are believed to be
(1991)
According to the universally accepted proton-neutron model of the nucleus, atomic nuclei are composed of protons and neutrons, which are collectively referred to as nucleons.
Question 144:
easyA nucleus of mass number 189 splits into two nuclei having mass number 125 and 64. The ratio of radius of two daughter nuclei respectively is :
(2022)
The radius of a nucleus is related to its mass number by $R = R_0 A^{1/3}$. The ratio of their radii is $R_1 / R_2 = (A_1 / A_2)^{1/3} = (125 / 64)^{1/3}$. This gives $R_1 / R_2 = 5 / 4$, so the ratio is $5 : 4$.
Question 145:
easyThe energy equivalent of 0.5 g of a substance is :
(2020)
Using Einstein's mass-energy equivalence principle $E = mc^2$. Substituting $m = 0.5 g = 0.5 \times 10^{-3} kg$ and $c = 3 \times 10^8 m/s$. $E = (0.5 \times 10^{-3}) \times (3 \times 10^8)^2 = 0.5 \times 10^{-3} \times 9 \times 10^{16} = 4.5 \times 10^{13} J$.
Question 146:
easyIf radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$, then the radius of $^{125}_{53}Te$ nucleus is nearly:
(2015)
Nuclear radius $R \propto A^{1/3}$. The ratio of radii is $R_{Te} / R_{Al} = (A_{Te} / A_{Al})^{1/3} = (125 / 27)^{1/3}$. This simplifies to $R_{Te} / R_{Al} = 5 / 3$, meaning $R_{Te} = \frac{5}{3} R_{Al}$.
Question 147:
easyIf the nuclear radius of $^{27}Al$ is 3.6 Fermi, the approximate nuclear radius of $^{64}Cu$ in Fermi is:
(2012 Pre)
The nuclear radius follows the relation $R \propto A^{1/3}$. Therefore, $R_{Cu} / R_{Al} = (64 / 27)^{1/3} = 4 / 3$. $R_{Cu} = (4/3) \times 3.6 = 4.8 Fermi$.
Question 148:
easyTwo nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be:
(2008)
Nuclear density is roughly constant for all nuclei and is independent of the mass number A. Therefore, regardless of their mass numbers, the ratio of their nuclear densities is $1 : 1$.
Question 149:
easyIf the nucleus $^{27}_{13}Al$ has nuclear radius of about 3.6 fm, then $^{125}_{52}Te$ would have its radius approximately as:
(2007)
Using the relation $R \propto A^{1/3}$, we get $R_{Te} = R_{Al} (A_{Te} / A_{Al})^{1/3}$. $R_{Te} = 3.6 \times (125 / 27)^{1/3} = 3.6 \times (5 / 3) = 6.0 fm$.
Question 150:
easyThe radius of germanium (Ge) nuclide is measured to be twice the radius of $^{9}_{4}Be$. The number of nucleons in Ge are:
(2006)
Given $R_{Ge} = 2 R_{Be}$, we can write $R_0 A_{Ge}^{1/3} = 2 R_0 (9)^{1/3}$. Canceling $R_0$ and cubing both sides gives $A_{Ge} = 2^3 \times 9 = 8 \times 9 = 72$. The number of nucleons is 72.