Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B1 Radiation Physics, National Examination
December 2013 — a three-hour open-book examination in which any
non-communicating calculator is permitted. The cover page states that all seven
questions must be attempted (no choose-N-of-M here) for a total of 100 points, and
invites the candidate to submit a written statement of any assumptions made where a
question is open to interpretation. This licence is used below in Question 2(c) (the
photon-production law assumed for the current change) and Question 5(a) (the numeric value
behind the news item's "five times the acceptable exposure" claim).
Reference texts. K. S. Krane, Introductory Nuclear Physics
(nuclear masses and binding energy, radioactive decay, fission); F. H. Attix,
Introduction to Radiological Physics and Radiation Dosimetry (X-ray production and
bremsstrahlung spectra, photon interactions — photoelectric effect, Compton
scattering, pair production, gamma detectors); J. R. Cember and T. E. Johnson,
Introduction to Health Physics, 5th ed. (dose equivalent, internal dosimetry and
effective half-life, shielding, fission-product hazards); J. E. Turner, Atoms,
Radiation, and Radiation Protection, 3rd ed. (radiation interactions with matter,
health-physics standards).
Given. Infrared wavelength band $\lambda = 0.7$–$300\ \mu$m;
sodium ionization energy $> 6$ eV.
Find. The corresponding frequency and photon-energy ranges, whether
infrared can ionize sodium, and how photon energy behaves under reflection versus
Compton scattering.
Approach. Convert the wavelength band directly with $f=c/\lambda$ and
$E=hc/\lambda$, compare the resulting maximum photon energy against sodium's ionization
threshold, then contrast the elastic (energy-conserving) nature of macroscopic reflection
with the inelastic, photon–electron nature of Compton scattering.
Part (a) — frequency range. $f = c/\lambda$, with the longer
wavelength giving the lower frequency:
$$f_{\text{low}} = \frac{3\times10^8}{300\times10^{-6}} = 1.0\times10^{12}\text{ Hz},
\qquad
f_{\text{high}} = \frac{3\times10^8}{0.7\times10^{-6}} = 4.29\times10^{14}\text{ Hz}$$
$$\boxed{f \approx 1\times10^{12}\text{ Hz to } 4.29\times10^{14}\text{ Hz (1 THz to 429 THz)}}$$
Part (b) — photon energy. $E=hc/\lambda$ evaluated at both band
edges (photon energy rises as wavelength shortens, opposite sense to part (a)):
$$E_{\text{low}} = \frac{hc}{300\ \mu\text{m}} = 4.14\times10^{-3}\text{ eV},
\qquad
E_{\text{high}} = \frac{hc}{0.7\ \mu\text{m}} = 1.77\text{ eV}$$
$$\boxed{E \approx 0.004\text{ eV to } 1.77\text{ eV across the infrared band}}$$
Part (c) — not ionizing for sodium. The most energetic infrared
photon in the band (shortest wavelength, 0.7 $\mu$m) carries only 1.77 eV, from part (b).
Since $1.77\text{ eV} < 6\text{ eV}$, no photon anywhere in the infrared band carries
enough energy to eject a bound electron from sodium in a single photon absorption —
$$\boxed{E_{\text{IR,max}} = 1.77\text{ eV} \ll E_{\text{ionize,Na}} > 6\text{ eV}}$$
which is exactly the physical definition of non-ionizing radiation for that atom: no single
photon can supply the energy the electron needs to escape.
Part (d) — energy under reflection. Reflection off a macroscopic
surface (mirror, polished metal) is an elastic, coherent wave phenomenon governed
by boundary conditions on the electromagnetic field: the photon's frequency, and therefore
its energy $E=hf$, is unchanged by an ordinary (stationary-mirror)
reflection.
Part (e) — energy under scattering. A gamma-ray photon undergoing
Compton scattering transfers part of its energy and momentum to the (effectively free)
electron it scatters from; the scattered photon's energy is therefore
lower than before the collision, following
$$E' = \frac{E}{1+\dfrac{E}{m_ec^2}(1-\cos\theta)} < E$$
Part (f) — is there a real difference? Yes. Optical
"reflection" is a coherent, elastic interaction with a macroscopic boundary in which the
wave picture dominates and photon energy is conserved (only direction changes). Gamma-ray
"scattering" (Compton scattering) is an inelastic, particle-like photon–electron
collision in which energy and momentum are genuinely exchanged with an individual electron,
so the photon loses energy. The two terms describe physically distinct processes; the
choice of word in the question is not interchangeable, and reflects the different photon
energy regimes (optical vs. gamma) in which each dominates.
