Correct option is (d)

As this is a reverse bias characteristic. It should be for Zener diode working as voltage regulator.

Correct option is (b)

The wavelength of light emitted by a Light Emitting Diode (LED) fabricated using a semiconducting material can be determined by the energy band gap of the material. The energy of the photon emitted, which corresponds to the band gap energy, is given by the equation:

$E=\frac{hc}{\lambda }$

Where:

• $E$ is the energy of the emitted photon (in joules when $h$ and $c$ are in SI units), corresponding to the band gap energy of the material.
• $h$ is Planck's constant ($6.626\times {10}^{-34}{\mathrm{m}}^2\mathrm{kg}/\mathrm{s}$).
• $c$ is the speed of light in vacuum ($3.0\times {10}^8\mathrm{m}/\mathrm{s}$).
• $\lambda$ is the wavelength of the emitted light (in meters).

However, since the energy band gap given is in electronvolts (eV), and we are looking for the wavelength in nanometers (nm), we can use the energy formula directly in terms of eV and then do the unit conversion conveniently. The conversion between energy (in eV) and wavelength (in nm) without needing to convert eV to Joules is facilitated by the equation:

$\lambda (\mathrm{nm})=\frac{1240}{E(\mathrm{eV})}$

Here, 1240 nm·eV is a conversion factor used for directly converting energy in eV to wavelength in nm.

Given the band gap energy of GaAs is $1.42\mathrm{eV}$, the wavelength ($\lambda$) of light emitted can be found as:

$\lambda =\frac{1240}{1.42}=873.24\mathrm{nm}$

Thus, the wavelength of light emitted from the LED is approximately 873.24 nm, which is closest to:

Option B: 875 nm.

Correct option is (d)

Pd across $R_s$

${\mathrm{V}}_1=8-5=3\mathrm{V}$

Current through the load resistor

$\mathrm{I}=\frac{5}{1\times {10}^3}=5\mathrm{mA}$

Maximum current through Zener diode

${\mathrm{I}}_{\mathrm{z}max.}=\frac{10}{5}=2\mathrm{mA}$

And minimum current through Zener diode

$\begin{array}{c}{\mathrm{I}}_{\mathrm{z}\text{ min. }}=0\\ \\ \therefore {\mathrm{I}}_{\mathrm{s}\text{ max. }}=5+2=7\mathrm{mA}\end{array}$

And ${\mathrm{R}}_{\mathrm{s}\text{ min }}=\frac{{\mathrm{V}}_1}{{\mathrm{I}}_{\mathrm{s}\text{ max }}}=\frac{3}{7}\mathrm{k}\mathrm{\Omega }$

Similarly ${\mathrm{I}}_{\mathrm{s}min.}=5\mathrm{mA}$

And ${\mathrm{R}}_{\mathrm{s}\text{ max. }}=\frac{{\mathrm{V}}_1}{{\mathrm{I}}_{\mathrm{s}\text{ min. }}}=\frac{3}{5}\mathrm{k}\mathrm{\Omega }$

$\therefore \frac{3}{7}\mathrm{k}\mathrm{\Omega }<{\mathrm{R}}_{\mathrm{s}}<\frac{3}{5}\mathrm{k}\mathrm{\Omega }$

Correct option is (b)

Zener is in breakdown region.

$\begin{array}{rl}& I_3=\frac{10}{500}=\frac{1}{50}\\ \\ & I_1=\frac{10}{200}=\frac{1}{20}\\ \\ & I_2=I_1-I_3\\ \\ & I_2=(\frac{1}{20}-\frac{1}{50})=\left(\frac{3}{100}\right)=30\mathrm{mA}\end{array}$

Correct option is (c)

${\mathrm{V}}_{\mathrm{z}}=3\mathrm{V}$

Let potential at $\mathrm{B}=0\mathrm{V}$

Potential at $E\left(V_E\right)=10V$

$\begin{array}{rl}& V_C=V_A=3V\\ & I_z+I_1=I\\ & I=\frac{10-3}{1000}=\frac{7}{1000}A\\ & I_1=\frac{3}{2000}A\end{array}$

Therefore $I_z=\frac{7-1.5}{1000}=5.5\mathrm{mA}$