NEET Practice Q & A

Hint

(d) Concept:
Relation between Current Density and Electric Field:
Current Density:
The amount of electric current travelling per unit cross-section area is called as current density and expressed in amperes per square meter.
Formula for Current Density is given as,
\(
\mathrm{J}=\mathrm{I} / \mathrm{A}
\)
Where,
I = current flowing through the conductor in Amperes
\(A=\) cross sectional area in \(\mathrm{m}^2\).
Consider a conductor of length I and area of cross-section A. Let its two ends be raised to potentials \(\mathrm{V}_1[latex] and [latex]\mathrm{V}_2\left(\mathrm{~V}_1>\mathrm{V}_2\right)\).
The electric current density is given by the formula:
\(
j=\sigma E=n e v_d
\)
Calculation:
Given:
Density of electrons, \(\mathrm{n}_{\mathrm{e}}=10^{19} \mathrm{~m}^{-3}\)
Charge of electron, \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\)
Mobility of electron, \(\mu_e=1.6 \mathrm{~m}^2 /(\mathrm{V} . \mathrm{s})\)
The electric current density is given by the formula:
\(
\mathrm{j}=\sigma \mathrm{E}=\mathrm{nev}_{\mathrm{d}}
\)
For intrinsic semiconductors (no impurities), the number of electrons will be equal to the number of holes.
\(
\sigma=\mathrm{ne} \frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{E}}=\mathrm{ne} \mu
\)
For N -type semiconductor, electrons are majority carriers.
Conductivity, \(\sigma \approx[latex] ne [latex]\mu\)
Resistivity, \(\rho=\frac{1}{\sigma}=\frac{1}{\mathrm{n}_{\mathrm{e}} \mathrm{e} \mu_{\mathrm{e}}} \dots(1)\)
By substituting the given values in the equation (1),
\(
\begin{aligned}
& \rho=\frac{1}{10^{19} \times 1.6 \times 10^{-19} \times 1.6} \\
& \rho=0.4 \Omega \mathrm{~m}
\end{aligned}
\)

Hint

(c) The temperature coefficient of resistance is equal to the change in resistance of the wire of resistance one ohm at \(0^{\circ} \mathrm{C}\) when the temperature changes by \(1^0 \mathrm{C}\).
That is denoted by \(\alpha\).
\(
\alpha=\frac{R_t-R_o}{R_o t}
\)
Where,
\(\alpha=\) temperature coefficient
Let’s consider a case,
If \(R_o=1\) and \(\mathrm{t}=1^0 \mathrm{C}\) then, \(\alpha=R_t-R_o\)
In general, for two different temperature and resistance, we can write the equation as,
\(
\alpha=\frac{R_2-R_1}{R_1 t-R_2 t}
\)
From this equation, we can say that as the temperature of a semiconductor increases, resistance decreases.
Then the value \(R_2-R_1\) becomes a negative value. Therefore the entire value becomes negative in the case of the semiconductor.
Thus, the temperature coefficient of resistance \((\alpha)\) of a semiconductor is always negative.

Hint

(b) Energy bands in solids are a consequence of (b) Pauli exclusion principle.

Explanation: The Pauli exclusion principle states that no two electrons in an atom can have the same set of quantum numbers, which leads to the formation of distinct energy levels when atoms come together to form a solid.

Why other options are incorrect:
(a) Ohm’s law:

Ohm’s law relates current, voltage, and resistance in a conductor. It doesn’t explain the formation of energy bands.
(c) Bohr’s theory:

Bohr’s theory describes the quantized energy levels of electrons in individual atoms, but it doesn’t explain how these levels merge into bands when atoms form a solid.
(d) Heisenberg’s uncertainty principle:

Heisenberg’s uncertainty principle deals with the limitations in simultaneously knowing the position and momentum of a particle. While it’s a fundamental principle in quantum mechanics, it doesn’t directly explain the formation of energy bands.

Hint

(b) Compare the change in current through the diode and the resistor when the voltage is increased.
Step 1: Initially, the readings of A1 and A2 are the same, meaning the current through the diode ( \(I_D\) ) and the current through the resistor ( \(I_R\) ) are equal:
\(I_D=I_R\)

Step 2: The current through the diode increases exponentially with voltage:
\(I_D=I_0\left(e^{\frac{c V}{n k T}}-1\right)\)
Where \(I_0\) is the reverse saturation current, \(\boldsymbol{e}\) is the electron charge, \(V\) is the voltage, \(n\) is the ideality factor, \(k\) is Boltzmann’s constant, and \(T\) is the temperature.
When the voltage increases, the exponential term dominates, and the current through the diode increases significantly.

Step3: The current through the resistor increases linearly with voltage according to Ohm’s law:
\(I_R=\frac{V}{R}\)
When the voltage increases, the current through the resistor also increases, but linearly.

Step4: Therefore, the reading of A1 (diode current) will be greater than the reading of A2 (resistor current).

Hint

(d) When an electron diffuses from \(n \rightarrow p\), it leaves behind an ionised donor on \(n\)-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from \(n \rightarrow p\), a layer of positive charge (or positive space-charge region) on n -side of the junction is developed.

Similarly, when a hole diffuses from \(\mathrm{p} \rightarrow \mathrm{n}\) due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its on diffusion free charges (Fig. D).

Hint

(c) The given frequency of the input \(=50 \mathrm{~Hz}\)
The fundamental frequency of the output ripples \(=2\) times the input frequency \(=2 \times 50=100 \mathrm{~Hz}\).

Hint

(a) To obtain a P-type silicon semiconductor, you need to dope pure silicon with a trivalent impurity like boron, aluminum, or gallium.

Explanation: When a trivalent impurity is added to silicon, it creates “holes” in the crystal lattice, allowing for positive charge carriers to move, which is characteristic of a P-type semiconductor.

Hint

(c)

\(
\begin{aligned}
&\text { By Kirchhoff’s node law, }\\
&I_T=I_Z+I_L
\end{aligned}
\)
\(
I_T=\frac{(100-40)}{3K}=20 \mathrm{~mA}
\)
\(
I_L=\frac{40}{6K}=\frac{20}{3} \mathrm{~mA}
\)
\(
I_Z=20-\frac{20}{3}= \frac{40}{3} \mathrm{~mA}
\)

Hint

(b) Diode D1 and D4 will be forward biased (D2 and D3 will be reverse biased) and replaced with 50 Ohm resistor each.

\(
\begin{aligned}
I & =\frac{V}{R_t} \\
I & =\frac{12 \mathrm{~V}}{300 \Omega} \\
I & =0.04 \mathrm{~A}
\end{aligned}
\)

Hint

(d) \(X=1, Y=0\)

Hint

(d)

\(
Y=A\bar{B}
\)

Hint

(d)

Calculate the energy of the photon with the given wavelength and equate it to the band gap energy. Then, solve for \(n\).

Calculate the energy of the photon:

The energy of the photon is given by:
\(E=\frac{h c}{\lambda}\)

Substituting the given values:
\(E=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{2480 \times 10^{-9} \mathrm{~m}}\)
\(E=\frac{19.878 \times 10^{-26}}{2480 \times 10^{-9}} \mathrm{~J}\)
\(E=8.015 \times 10^{-20} \mathrm{~J}\)

Convert the energy to electron volts (eV)
Using the conversion factor \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\) :
\(E=\frac{8.015 \times 10^{-20} \mathrm{~J}}{1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
\(E=0.500 \mathrm{eV}\)

Equate the energy to the band gap and solve for \(\boldsymbol{n}\)
The band gap is given as \(n \times 10^{-2} \mathrm{eV}\) :
\(0.500 \mathrm{eV}=n \times 10^{-2} \mathrm{eV}\)
\(n=\frac{0.500}{10^{-2}}\)
\(n=50\)

Hint

(b) Step 1: Express the ratio of currents in terms of concentrations and drift velocities.
The electron current is \(I_e=n_e e A v_e\).
The hole current is \(I_h=n_h e A v_h\).
The ratio of currents is \(\frac{I_e}{I_h}=\frac{n_e e A v_e}{n_h e A v_h}\).
Simplify the ratio: \(\frac{I_e}{I_h}=\frac{n_e v_e}{n_h v_h}\).
Step 2: Solve for the ratio of drift velocities.
Rearrange the equation from Step 1: \(\frac{v_e}{v_h}=\frac{I_e}{I_h} \cdot \frac{n_h}{n_e}\).
Substitute the given ratios: \(\frac{v_e}{v_h}=\frac{7}{4} \cdot \frac{5}{7}\).
Simplify: \(\frac{v_e}{v_h}=\frac{5}{4}\).

Step 3: Find \(n\).
We are given that \(\frac{v_e}{v_h}=\frac{n}{4}\).
From Step 2, we found that \(\frac{v_e}{v_h}=\frac{5}{4}\).
Equate the two expressions: \(\frac{n}{4}=\frac{5}{4}\).
Solve for \(n: n=5\).
Solution
The value of \(n\) is 5 .