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Q10 PE
Expert-verifiedThe 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: \(931.53\,{\rm{MeV}}\).
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.
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: \(931.53\,{\rm{MeV}}\).
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