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Nuclei - CBSE Class 12th NCERT Solution

Sep 2, 2013 05:36 IST

    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 12th 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

     To get the complete study material for this chapter, Click Here 

    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

    To Get the Complete Solved Chapter, Click Here
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