RankLab
Back to Questions
PhysicsMediumMCQ2020 · 04 Sep Shift 2

Q19.Find the Binding energy per nucleon for 12050Sn. Mass of proton mp = 1. 00783 U , mass of neutron mn = 1. 00867 U and mass of tin nucleus msn = 119. 902199 U . (take 1U = 931 MeV ) (1) 7 .5 MeV (2) 9. 0MeV (3) 8.0MeV (4) 8 .5 MeV

What This Question Tests

This question requires calculating the mass defect, then the total binding energy using the mass-energy equivalence, and finally the binding energy per nucleon.

Concepts Tested

Mass defectBinding energyBinding energy per nucleonMass-energy equivalence

Formulas Used

Δm = (Zmp + (A-Z)mn) - M_nucleus

BE = Δm × 931.5 MeV/U

BE per nucleon = BE / A

📚 NCERT Sections This Tests

13.2Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of

Physics Class 12 · Chapter 13

85% match

13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u

13.3A Given Coin Has A Mass Of 3.0 G. Calculate The Nuclear Energy That

Physics Class 12 · Chapter 13

82% match

13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).

13.5The Q Value Of A Nuclear Reaction A + B ® C + D Is Defined By

Physics Class 12 · Chapter 13

81% match

13.5 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. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u