Q13 PE
Question
The detail observable using a probe is limited by its wavelength. Calculate the energy of a \({\rm{\gamma }}\) -ray photon that has a wavelength of\(1 \times {10^{ - 16}}\,{\rm{m}}\), small enough to detect details about one-tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe.
Step-by-Step Solution
VerifiedThe energy of \({\rm{\gamma }}\)ray photon is obtained as: \(E = 1241\,{\rm{MeV}}\).
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 energy of electromagnetic radiation is obtained using:
\(E = \frac{{hc}}{\lambda }.................(1)\)
The value of energy is \({\rm{e}}\).
The value of plancks is \({\rm{h}}\).
We have the value of speed of light as \({\rm{c}}\) which is constant.
The value of wavelength is \({\rm{\lambda }}\).
We also have: \(\lambda = 1x{10^{ - 16}}\,{\rm{m}}\).
So, energy is then obtained as:
\(\begin{align}E &= \frac{{(6.62 \times {{10}^{ - 34}}\,Js)(3.00 \times {{10}^8}\,m/s)}}{{(1 \times {{10}^{ - 16}}\,m)}}\\ &= 1.99 \times {10^{ - 9}}\,J\\ &= 124\,MeV....................(2)\end{align}\)
Therefore, the energy is: \(E = 1241\,{\rm{MeV}}\).