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.
Show Answer
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 $
