Question #1549b

1 Answer
A08
Mar 24, 2017

We know that one atomic mass unit is defined as 1/12th the mass of an atom of Carbon-12 isotope.
As such #1" amu"=1.6606xx10^-27" kg"#
Velocity of light #c# in vacuum is #=2.9979xx10^8" ms"^-1#

Using the Einstein's mass energy equivalence relation we get
#E=mc^2#
#E=1.6606xx10^-27xx(2.9979xx10^8)^2" J"#

We also know that #1" eV"=1" electronic charge in C"xx1 " V"#
#=1.60218xx10^-19" J"#

Therefore, we get
#E=(1.6606xx10^-27xx(2.9979xx10^8)^2)/(1.60218xx10^-19)" eV"#
#=>E=931.5" MeV"#, rounded to one decimal place

Related questions
See all questions in Energy
Impact of this question
3834 views around the world
You can reuse this answer
Creative Commons License