Question 331:
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 332:
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 333:
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 334:
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 335:
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 336:
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.
Question 337:
easyThe nuclei of which one of the following pairs of nuclei are isotones?
(2005)
Isotones are nuclei with the same number of neutrons ($N = A - Z$). For $_{34}Se^{74}$, $N = 74 - 34 = 40$. For $_{31}Ga^{71}$, $N = 71 - 31 = 40$. Since both have 40 neutrons, they are isotones.
Question 338:
easyIn the nucleus of $ _{11}Na^{23} $, the number of protons, neutrons and electrons are
(1991)
For a nucleus represented as $ _{Z}X^{A} $, the number of protons is $ Z = 11 $. The number of neutrons is $ A - Z = 23 - 11 = 12 $. Since it is a nucleus, there are no electrons present, so electrons = 0.
Question 339:
easyThe nuclei $ _{6}C^{13} $ and $ _{7}N^{14} $ can be described as
(1990)
The number of neutrons in $ _{6}C^{13} $ is $ 13 - 6 = 7 $. The number of neutrons in $ _{7}N^{14} $ is $ 14 - 7 = 7 $. Nuclei with the same number of neutrons are called isotones.
Question 340:
easyA nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:
(2021)
Initial binding energy = $ 240 \times 7.6 $ MeV. Final binding energy of the two fragments = $ 2 \times (120 \times 8.5) = 240 \times 8.5 $ MeV. Total gain in binding energy = $ 240 \times (8.5 - 7.6) = 240 \times 0.9 = 216 $ MeV.