In the given circuit, the voltage across load resistance ( R _ L )...

Both diodes are in forward bias. In forward bias, potential drop across $Ge$ and $Si$ will be $0 . 3V \text{ and } 0 . 7V$ respectively.

Therefore, current in the circuit $i = \dfrac{15 - 0 . 3 - 0 . 7}{4 \times 1000} = 3 . 5mA$

Required potential drop across ${R}_{L}$, 
${V}_{L} = i{R}_{L} = \left(3 . 5 \times {10}^{- 3}\right) \times \left(2 . 5 \times 1000\right)V$

$= 8 . 75V$