The laptop PCs,modern electronic watches,and calculators use the following for display:

Statement $-1$: $A$ photodiode operates in reverse bias.
Statement $-2$: The fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.

The figure shows $2$ $LEDs$ that can be used as a polarity detector. Applying a positive source voltage results in a green light,while negative supplies result in a red light. Packages of such combinations are commercially available. Find the resistor $R$ to ensure a current of $20 \ mA$ through the $ON$ diode for the configuration. Both diodes have a reverse breakdown voltage of $3 \ V$ and an average turn-on voltage of $2 \ V$. The source voltage is $8 \ V$. Find $R$ in $\Omega$.

For three circuit elements $A, B$ and $C$,following observations were made in lab:
$(a)$ element $A$ cannot conduct $d.c.$ but allows $a.c.$ through it.
$(b)$ element $B$ conducts both $AC$ and $DC$ but in one direction only.
$(c)$ Element $C$,when connected across $DC$,conducts only for one polarity and emits light whose intensity first increases and then decreases with increase in applied voltage.
The elements $A, B$ and $C$ are respectively identified as

$LED$ is constructed from $GaAsP$ semiconducting material. The energy gap of this $LED$ is $1.9\, eV$. Calculate the wavelength of light emitted and its colour.
$[h = 6.63 \times 10^{-34} \, Js$ and $c = 3 \times 10^{8} \, ms^{-1}]$

For $LED$s to emit light in the visible region of the electromagnetic spectrum,they should have an energy band gap in the range of: