Problem 6.33 — K-shell Binding Energy from Absorption Edge

Problem Statement

Solve the work-energy problem: Solve the nuclear physics problem: The K-absorption edge of silver is at $\lambda_K = 0.0485$ nm. Find the K-shell binding energy. $$E_K = hc/\lambda_K = 1240 \text{ eV·nm}/0.0485 \text{ nm} = 25567 \text{ eV} \approx 25.6 \text{ keV}$$ Answer: $E_K \approx 25.6$ keV Nuclide symbol, atomic number $Z

Given Information

• Mass $m$, velocity $v$, height $h$, or other given quantities
• Any forces doing work (conservative or non-conservative) as specified

Physical Concepts & Formulas

The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta KE$. For conservative forces (gravity, spring, electric), a potential energy function $U$ exists such that $W = -\Delta U$, and the total mechanical energy $E = KE + U$ is conserved. Non-conservative forces (friction, air drag) remove mechanical energy, converting it to thermal energy. The power delivered is $P = dW/dt = \vec{F}\cdot\vec{v}$.

• $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$ — work done by constant force
• $KE = \frac{1}{2}mv^2$ — kinetic energy
• $U_g = mgh$ — gravitational PE (near Earth’s surface)
• $U_s = \frac{1}{2}kx^2$ — elastic PE
• $W_{\text{net}} = \Delta KE = KE_f – KE_i$ — work-energy theorem
• $E_i = E_f$ (when only conservative forces act)

Step-by-Step Solution

Step 1 — Identify all forces and whether they are conservative.

Step 2 — Apply conservation of energy (if no friction):

$$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$$

Step 3 — If friction acts:

$$E_i – W_{\text{friction}} = E_f \implies \frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f + f_k d$$

Step 4 — Solve for the unknown (usually $v_f$ or $d$).

Worked Calculation

Substituting all values with units:

Ball of mass $m = 0.5\,\text{kg}$ dropped from $h = 10\,\text{m}$:

$$v_f = \sqrt{2gh} = \sqrt{2\times9.8\times10} = \sqrt{196} = 14\,\text{m/s}$$

Answer

$$\boxed{v_f = \sqrt{2g h}}$$

Physical Interpretation

A 14 m/s impact speed from a 10 m fall corresponds to about 50 km/h — enough to be seriously dangerous. This underscores why falling from height is hazardous. The conversion of gravitational PE to KE is 100% efficient in the absence of air resistance — every joule of lost PE appears as gained KE.