Question 91:
easyThe mass of a $ _3^7Li $ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of $ _3^7Li $ nucleus is nearly:
(2010 Pre)
Mass defect $ \Delta m = 0.042 $ u. Total binding energy $ = \Delta m \times 931.5 $ MeV = $ 0.042 \times 931.5 = 39.123 $ MeV. Binding energy per nucleon = $ 39.123 / 7 \approx 5.6 $ MeV.
Question 92:
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 93:
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 94:
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.
Question 95:
easyThe power obtained in a reactor using $ U^{235} $ disintegration is 1000 kW. The mass decay of $ U^{235} $ per hour is:
(2011 Pre)
Power $ P = 1000 $ kW $ = 10^6 $ J/s. Energy produced in one hour $ E = 10^6 \times 3600 = 3.6 \times 10^9 $ J. Using $ E = mc^2 $, mass decay $ m = E / c^2 = (3.6 \times 10^9) / (3 \times 10^8)^2 = 4 \times 10^{-8} $ kg = 40 micrograms.
Question 96:
easyThe binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
(2010 Mains)
Binding energy of two deuterium nuclei = $ 2 \times (2 \times 1.1) = 4.4 $ MeV. Binding energy of the helium nucleus = $ 4 \times 7.0 = 28.0 $ MeV. Energy released = Final BE - Initial BE = $ 28.0 - 4.4 = 23.6 $ MeV.
Question 97:
easyIf $ M(A, Z) $, $ M_p $ and $ M_n $ denote the masses of the nucleus $ ^A_Z X $, proton and neutron respectively in units of u (1 u = 931.5 $ MeV/c^2 $) and BE represents its binding energy in MeV, then:
(2008)
The binding energy is given by $ B.E. = \Delta m c^2 = [Z M_p + (A-Z) M_n - M(A, Z)] c^2 $. Rearranging this gives the nuclear mass $ M(A,Z) = Z M_p + (A-Z) M_n - B.E./c^2 $.
Question 98:
easyA nucleus $ ^A_Z X $ has mass represented by $ M(A, Z) $. If $ M_p $ and $ M_n $ denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then:
(2007)
Binding energy is the energy equivalent of the mass defect. The mass defect is $ \Delta m = [Z M_p + (A-Z) M_n - M(A, Z)] $. Thus, the binding energy is $ B.E. = \Delta m c^2 = [Z M_p + (A-Z) M_n - M(A, Z)]c^2 $.
Question 99:
easy$ M_p $ denotes the mass of a proton and $ M_n $ that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is velocity of light):
(2004)
The binding energy B is given by $ B = [Z M_p + N M_n - M(N, Z)] c^2 $. Rearranging for the mass of the nucleus gives $ M(N, Z) = Z M_p + N M_n - B/c^2 $.
Question 100:
easyThe mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of $ _2^4He $ is (Given: helium nucleus mass $ \approx 4.0015 $ u)
(2003)
Mass of constituents = $ 2(1.0073) + 2(1.0087) = 4.0320 $ u. Mass defect $ \Delta m = 4.0320 - 4.0015 = 0.0305 $ u. Binding energy $ = 0.0305 \times 931.5 $ MeV $ \approx 28.4 $ MeV.