In a light emitting diode $(LED)$,light is emitted due to:

Three photodiodes $D_1, D_2,$ and $D_3$ are made of semiconductors having band gaps of $2.5 \ eV, 2 \ eV,$ and $3 \ eV,$ respectively. Which ones will be able to detect light of wavelength $6000 \ \mathring{A}$?

$A$ $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.8 \; eV$. Can it detect a wavelength of $6000 \; nm$?

Given below are two statements: one is labeled as Assertion $A$ and the other is labeled as Reason $R$.
Assertion $A:$ Photodiodes are used in forward bias usually for measuring the light intensity.
Reason $R:$ For a $p-n$ junction diode, at applied voltage $V$, the current in the forward bias is more than the current in the reverse bias for $|V_z| > |V| \geq V_0$, where $V_0$ is the threshold voltage and $V_z$ is the breakdown voltage.
In the light of the above statements, choose the correct answer from the options given below.

An $LED$ is manufactured using zinc selenide; then it emits:

$A$ light emitting diode $(LED)$ has a voltage drop of $2\,V$ across it and passes a current of $10\,mA$ when it operates with a $6\,V$ battery through a limiting resistor $R$. The value of $R$ is $......\,k\Omega$.