For the circuit shown below, the current through the Zener diode is:

Test potential across load resistance ($$R_L = 10\text{ k}\Omega$$) assuming Zener diode is disconnected:

$$V_L = V_s \left(\frac{R_L}{R_s + R_L}\right)$$

$$V_L = 120 \left(\frac{10}{5 + 10}\right) = 120 \times \frac{10}{15} = 80\text{ V}$$

Since $$V_L > V_Z$$ ($$80\text{ V} > 50\text{ V}$$), the Zener diode operates in the breakdown region: $$V_L = V_Z = 50\text{ V}$$

$$I_s = \frac{V_s - V_Z}{R_s} = \frac{120 - 50}{5\text{ k}\Omega} = \frac{70}{5 \times 10^3} = 14\text{ mA}$$

$$I_L = \frac{V_Z}{R_L} = \frac{50}{10\text{ k}\Omega} = 5\text{ mA}$$

$$I_Z = I_s - I_L = 14 - 5 = 9\text{ mA}$$