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Q10 PE

Expert-verified
Found in: Page 1149

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

The unified atomic mass unit is defined to be$$1\,{\rm{u}} = 1.6605 \times {10^{ - 27}}\,{\rm{kg}}$$. Verify that this amount of mass converted to energy yields $$931.5\,{\rm{MeV}}$$. Note that you must use four-digit or better values for $${\rm{c}}$$ and $${\rm{|}}{{\rm{q}}_{\rm{e}}}{\rm{|}}$$.

The energy in the value of MeV is obtained as: $$931.53\,{\rm{MeV}}$$.

See the step by step solution

Step by Step Solution

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Step 2: Evaluating the energy

The mass energy is:

$$E = m{c^2}$$

The value of: $$m = 1.6605 \times {10^{ - 27}}\,{\rm{kg}}$$.

The value of speed of light is: $$c = 2.9979 \times {10^8}\,{\rm{m/s}}$$.

The magnitude of electrons charge is: $$|{q_e}| = 1.6021 \times {10^{ - 19}}\,{\rm{C}}$$.

So , the energy is obtained as:

\begin{align}E &= m{c^2}\\ &= (1.6605 \times {10^{ - 27}}\,kg){(2.9979 \times {10^8}\,m/s)^2}\\ &= 1.4924 \times {10^{ - 10}}\,J\end{align}

The energy when converted in the value of eV is:

\begin{align}E &= \frac{{1.4924 \times {{10}^{ - 10}}\,J}}{{1.6021 \times {{10}^{ - 13}}\,J/MeV}}\\ &= 931.53\,MeV\end{align}

Therefore, the energy is: $$931.53\,{\rm{MeV}}$$.