Production and use of X-rays — practice questions

1. The maximum energy of a photon ($E_{max}$) is equal to the kinetic energy of an electron accelerated through the potential difference $V_a$. $E_{max} = eV_a$
2. Calculate this energy in Joules: $V_a = 80 \text{ kV} = 80 \times 10^3 \text{ V}$ $E_{max} = (1.60 \times 10^{-19} \text{ C}) \times (80 \times 10^3 \text{ V})$ $E_{max} = 1.28 \times 10^{-14} \text{ J}$
3. The energy of a photon is related to its wavelength by $E = hc/\lambda$. The minimum wavelength corresponds to the maximum energy. $E_{max} = hc/\lambda_{min}$
4. Rearrange for $\lambda_{min}$ and substitute the values: $\lambda_{min} = hc/E_{max}$ $\lambda_{min} = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.00 \times 10^8 \text{ m/s})}{1.28 \times 10^{-14} \text{ J}}$ $\lambda_{min} = 1.5539 \times 10^{-11} \text{ m}$

Answer: The minimum wavelength of the X-rays produced is $1.55 \times 10^{-11}$ m (to 3 s.f.).

1. Identify the known values: Thickness, $x = 2.5$ cm Linear attenuation coefficient, $\mu = 0.21$ cm⁻¹
2. State the attenuation formula: $I = I_0 e^{-\mu x}$
3. To find the fraction of intensity transmitted, we calculate the ratio $I/I_0$. $\frac{I}{I_0} = e^{-\mu x}$
4. Substitute the values into the formula: $\frac{I}{I_0} = e^{-(0.21 \text{ cm}^{-1}) \times (2.5 \text{ cm})}$
5. Calculate the exponent: $-(0.21 \times 2.5) = -0.525$
6. Calculate the ratio: $\frac{I}{I_0} = e^{-0.525} = 0.59155...$
7. Convert the ratio to a percentage: $0.59155 \times 100\% = 59.155\%$

Answer: Approximately 59.2% of the initial X-ray intensity is transmitted through the muscle.