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

(A) $ 5mA $
(B) $ zero $
(C) $ 14mA $
(D) $ 9mA $

Hint
Zener diode is a reverse-biased P-N junction diode made of heavily doped semiconductors and operated in the breakdown region. A Zener diode that works in the breakdown region can be used as a voltage regulator. When a reverse bias is applied the zener diode will breakdown. This voltage is called the Zener breakdown voltage.
 $ I = \dfrac{V}{R} $ (Where $ I $ stands for the current through the circuit, $ V $ stands for the potential difference applied to the circuit, $ R $ stands for the resistance of the circuit)

Complete step by step answer
Let us assume that the zener diode did not breakdown,
The total resistance of the circuit is, $ R = 10000 + 5000\Omega = 15000\Omega $
The applied voltage is given by, $ V = 120V $
To find the current through the $ 5k\Omega $ resistor, we have to find the potential drop across the resistor.
The voltage across the $ 5k\Omega $ resistor is, $ {V_1} = 120 - 50 = 70V $
The voltage drop across the $ 10k\Omega $ resistor is required to find the current across the resistor
The voltage across the $ 10k\Omega $ resistor will be, $ {V_2} = 50V $
The current through the $ 10k\Omega $ resistor will be,
The current through the zener diode can be written as,
 $\Rightarrow I = 14 - 5 = 9mA $
The answer is:Option (D): $ 9mA $ .

Note
The zener diode maintains a constant voltage across the load even when the load current or the input voltage changes. Due to heavy doping, the width of the depletion layer is very narrow. When a reverse bias is applied, a very strong electric field is developed across the depletion layer. This breaks the covalent bonds and an extremely large number of electrons and holes are produced. This is called the Zener breakdown. This gives rise to a reverse saturation current called the Zener current. Zener current is independent of the applied voltage.