Atoms And Nuclei Question 335

Question: Calculate binding energy of $ _{92}U^{238} $ . Given $ MU^{238}=238.050783amu $ $ m_n=1.008665amu $ and $ m_p=1.007825amu $

Options:

A) 801.7MeV

B) 18.7 MeV

C) 0.7 MeV

D) 1801.7 MeV.

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Answer:

Correct Answer: D

Solution:

  • Mass defect $ \Delta m=[Zm_{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 $



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