Nuclei - CBSE Class 12th NCERT Solution

Get, detailed solutions to the questions of the chapter Nuclei from NCERT textbooks. The objective is to helping students regarding the pattern of answering the question as per the cbse latest marking scheme. Cbse.jagranjosh.com provided you NCERT solutions for classes 12^th math and science subjects.

Some questions of this chapter are given here.

Q. Obtain the binding energy (in MeV) of a nitrogen nucleus (14 )7N,
given m (14 )7N =14.00307 u

Answer

Binding energy of 7N14 = ΔMc2
Mass defect = [mass of 7 protons + mass of 7 neutrons – actual mass of N]
                    = [14.11543 – 14.00307]
                    = 0.11236 u
Binding energy = 0.11236 u x 931.5 MeV / u
                         = 104.66 MeV

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Q. 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 63 29Cu atoms (of mass 62.92960 u).

Answer

Given:
Mass of coin = 0.003 kg
Mass of 29Cu63 = 62.9296 u
                            = 1.045261 x 10-25 kg

No. of Cu atoms in given sample
                         D = 0.003 / 1.045261 x 10-25
                             = 2.8701 x 1022 atoms

Mass defect of 1 Cu atom = 0.591935

BE of 1 Cu atom = 0.591935 x 931.5 = 551.387 MeV
So,
   BE of  D atoms = 2.8701 x 1022 x 551.387
                             = 1.583 x 1025 MeV
                             = 2.53 x 1012 J

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