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"1 amu=1.6606×10−27 kg
Velocity of light cc in vacuum is =2.9979xx10^8" ms"^-1=2.9979×108 ms−1

Using the Einstein's mass energy equivalence relation we get
E=mc^2E=mc2
E=1.6606xx10^-27xx(2.9979xx10^8)^2" J"E=1.6606×10−27×(2.9979×108)2 J

We also know that 1" eV"=1" electronic charge in C"xx1 " V"1 eV=1 electronic charge in C×1 V
=1.60218xx10^-19" J"=1.60218×10−19 J

Therefore, we get
E=(1.6606xx10^-27xx(2.9979xx10^8)^2)/(1.60218xx10^-19)" eV"E=1.6606×10−27×(2.9979×108)21.60218×10−19 eV
=>E=931.5" MeV"⇒E=931.5 MeV, rounded to one decimal place

Related questions
See all questions in Energy
Impact of this question
4054 views around the world
You can reuse this answer
Creative Commons License