Atoms And Nuclei Question 257
Question: The energy in MeV is released due to transformation of 1 kg mass completely into energy $ (c=3\times 10^{8}m/s) $ [Pb. PMT 2003]
Options:
A) $ 7.625\times 10,MeV $
B) $ 10.5\times 10^{29},MeV $
C) $ 2.8\times {{10}^{-28}},MeV $
D) $ 5.625\times 10^{29},MeV $
Show Answer
Answer:
Correct Answer: D
Solution:
$ E=\Delta mc^{2}=1\times {{(3\times 10^{8})}^{2}}=9\times 10^{10}J $
$ \Rightarrow E=\frac{9\times 10^{16}}{1.6\times {{10}^{-19}}}=5.625\times 10^{35}eV $ $ =5.625\times 10^{29}MeV. $
