For the circuit shown here, the current through the Zener diode is

Assume Zener diode does not undergo breakdown. Current in the circuit,

$I=\frac{120}{15000}=8\times {10}^{-3} \mathrm{A}=8 \mathrm{mA}$

So voltage drop across diode $=\left(10\mathrm{k}\Omega \right)(8 \mathrm{mA})=80 \mathrm{V}>50 \mathrm{V}$ Hence, the diode undergoes breakdown.

Current in $R_1,I_1=\frac{70}{5000}=14 \mathrm{mA}$ Current in  $R_2,I_2=\frac{50}{10000}=5 \mathrm{mA}$

Current through Zener diode $=14-5=9 \mathrm{mA}$