2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Nuclear Physics And Radioactivity

JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    Calculate binding energy of \[_{92}{{U}^{238}}\]. Given \[M({{U}^{238}})=238.050783\,amu\]\[{{m}_{n}}=1.008665\,amu\] and \[{{m}_{p}}=1.007825\,amu\]

    A)  801.7MeV

    B)  18.7 MeV

    C)  0.7 MeV

    D)  1801.7 MeV.

    Correct Answer: D

    Solution :

    [d] Mass defect \[\Delta m=[Z{{m}_{p}}+(A-Z){{m}_{n}}]-M({{U}^{238}})\] \[=[92\times 1.007825\,\,(238-92)\times 1.008665]-\] \[238.050783\] \[\Delta m=1.93421\,amu\] \[BE=1.93421\times 931.5\,MeV=1801.7\,MeV\]


You need to login to perform this action.
You will be redirected in 3 sec spinner