Unit 20: Medical Physics — Numericals

Given Eᵢ = −3.5 × 10⁻¹⁹ J, E_f = −1.20 × 10⁻¹⁸ J, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹.

Energy released on the transition:
ΔE = |E_f − Eᵢ| = |−1.20 × 10⁻¹⁸ J − (−3.5 × 10⁻¹⁹ J)| = |−8.5 × 10⁻¹⁹ J| = 8.5 × 10⁻¹⁹ J

This energy is carried away by the emitted photon: ΔE = hc/λ, so
λ = hc/ΔE = (6.63 × 10⁻³⁴ J s)(3 × 10⁸ m s⁻¹) / (8.5 × 10⁻¹⁹ J)
λ = (1.989 × 10⁻²⁵ J m) / (8.5 × 10⁻¹⁹ J)
λ = 2.34 × 10⁻⁷ m = 234 nm

Result λ ≈ 234 nm, matching the printed answer.

Given V = 3000 V, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹.

If an electron loses all its kinetic energy in a single collision, the entire kinetic energy appears as one X-ray photon of maximum energy (minimum wavelength):
(K.E.)max = eV = hfmax = hc/λmin

So λmin = hc / (eV)

eV = (1.6 × 10⁻¹⁹ C)(3000 V) = 4.8 × 10⁻¹⁶ J

λmin = (6.63 × 10⁻³⁴ J s)(3 × 10⁸ m s⁻¹) / (4.8 × 10⁻¹⁶ J)
λmin = (1.989 × 10⁻²⁵ J m) / (4.8 × 10⁻¹⁶ J)
λmin = 4.14 × 10⁻¹⁰ m

Result λmin ≈ 4.14 × 10⁻¹⁰ m, matching the printed answer.

Given λ = 1.377 × 10⁻¹⁰ m, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹, e = 1.6 × 10⁻¹⁹ C.

The Kα photon energy equals the energy difference between the L- and K-shell levels:
ΔE = hc/λ = (6.63 × 10⁻³⁴ J s)(3 × 10⁸ m s⁻¹) / (1.377 × 10⁻¹⁰ m)
ΔE = (1.989 × 10⁻²⁵ J m) / (1.377 × 10⁻¹⁰ m)
ΔE = 1.444 × 10⁻¹⁵ J

Converting to electron-volts ΔE = (1.444 × 10⁻¹⁵ J) / (1.6 × 10⁻¹⁹ J/eV) ≈ 9028 eV ≈ 9.03 keV

Result ΔE ≈ 9.03 keV, matching the printed answer.

Given V = 40 kV = 4 × 10⁴ V, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹.

The shortest bremsstrahlung wavelength occurs when an electron loses all its kinetic energy in a single collision:
(K.E.)max = eV = hc/λmin ⟹ λmin = hc/(eV)

eV = (1.6 × 10⁻¹⁹ C)(4 × 10⁴ V) = 6.4 × 10⁻¹⁵ J

λmin = (6.63 × 10⁻³⁴ J s)(3 × 10⁸ m s⁻¹) / (6.4 × 10⁻¹⁵ J)
λmin = (1.989 × 10⁻²⁵ J m) / (6.4 × 10⁻¹⁵ J)
λmin ≈ 3.11 × 10⁻¹¹ m = 0.31 × 10⁻¹⁰ m

Result λmin ≈ 0.31 × 10⁻¹⁰ m, matching the printed answer.