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Uttarakhand PMT Uttarakhand PMT Solved Paper-2008

  • question_answer
    The mass of helium atom of mass number 4 is 4.0026 amu, while that of the neutron and proton are 1.0087 and 1.0078 respectively on the same scale. Hence, the nuclear binding energy per nucleon in the .helium atom is nearly

    A)  5MeV     

    B)  7MeV

    C)  10MeV      

    D)  14MeV

    Correct Answer: B

    Solution :

     He atom has \[2p+2n\] Hence, \[\Delta m=(2\times 1.0078+2\times 1.0087)-4.0026\] \[=0.0304amu\] \[\therefore \] energy released\[=0.0304\times 931.5MeV\] \[=28.3\text{ }MeV\] Binding energy per nucleon\[=\frac{28.3}{4}=7\,MeV\] (approximately)


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