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Q10 PE

College Physics (Urone)
Found in: Page 1149

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Short Answer

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

Step 1: Define Radioactivity

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: \(931.53\,{\rm{MeV}}\).


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