In a model of an atom, an electron goes round a circle of radius 2.5 × 10⁻¹⁰ m at a speed of 1.4 ×…

Step 1. The period. $T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 2.5 \times 10^{-10}\ \text{m}}{1.4 \times 10^{6}\ \text{m/s}} = 1.12 \times 10^{-15}\ \text{s}$.

Step 2. The current. $I = \dfrac{e}{T} = \dfrac{1.6 \times 10^{-19}\ \text{C}}{1.12 \times 10^{-15}\ \text{s}} = 1.43 \times 10^{-4}\ \text{A} = 0.143\ \text{mA}$.

Step 3. The moment. $m = I\pi r^2 = 1.43 \times 10^{-4}\ \text{A} \times \pi \times (2.5 \times 10^{-10}\ \text{m})^2 = 2.80 \times 10^{-23}\ \text{A m}^2$.

Step 4. The angular momentum. $L = m_evr = 9.1 \times 10^{-31}\ \text{kg} \times 1.4 \times 10^{6}\ \text{m/s} \times 2.5 \times 10^{-10}\ \text{m} = 3.19 \times 10^{-34}\ \text{kg m}^2/\text{s}$.

Step 5. The ratio. $\dfrac{m}{L} = \dfrac{2.80 \times 10^{-23}}{3.19 \times 10^{-34}} = 8.79 \times 10^{10}\ \text{C/kg}$.