Q.Hole is

Concept & Intuition

The question asks for current per unit area, which is the current density $J$. Instead of first finding total current and then dividing by area, we can use the microscopic form of Ohm’s law:

where $\sigma$ is conductivity and $E$ is the electric field inside the wire. This is much more direct because we already know the potential difference and length, so the electric field is simply $E = V/L$. No need to involve resistance, resistivity, or cross-sectional area at all.

Step-by-step solution

1. Identify the given quantities

   • Length of wire: $L = 100\ \text{cm} = 1.00\ \text{m}$
   • Potential difference: $V = 2\ \text{V}$
   • Conductivity: $\sigma = 5.95 \times 10^7\ \mathrm{S\,m^{-1}}$ (Siemens per metre, equivalent to $\Omega^{-1}\mathrm{m}^{-1}$)

2. Find the electric field inside the wire

   For a uniform wire, the electric field is constant and given by

3. Apply the microscopic Ohm’s law Current density $J$ is the product of conductivity and electric field: $$J = \sigma E = (5.95 \times 10^7) \times 2 …$$

Concept. The vector (microscopic) form of Ohm's law relates current density to the applied electric field through the material's conductivity: $\vec{J} = \sigma\vec{E} = \vec{E}/\rho$. The resistivity comes from the sample's dimensions via $R = \rho l / A$.

Step 1 — find the resistivity.

Step 2 — apply $J = E/\rho$. …

In a Common Emitter (CE) transistor amplifier, a small change in the input base-emitter voltage ($\Delta V_{BE}$) causes a significant change in the collector current ($\Delta I_C$). This change in collector current then flows through a load resistance ($R_L$) connected in the collector circuit, producing a large change in the output voltage ($\Delta V_{CE}$ or $\Delta V_{out}$). The voltage gain ($A_v$) of the amplifier is a measure of how much the amplifier increases the voltage of the input signal, defined as the ratio of the change in output voltage to the change in input voltage.

The core idea is that the transistor acts as a current amplifier, where a small base current change (driven by $\Delta V_{BE}$) leads to a larger collector current change. This amplified current change, when passed through a load resistor, converts back into a significantly amplified voltage change at the output.

Here's how to calculate the voltage gain:

1. Identify the input voltage change:

   The problem states that a signal of $25~\text{mV}$ is added to the base-emitter voltage. This is our input voltage change, $\Delta V_{in}$.

   $\Delta V_{in} = \Delta V_{BE} = 25~\text{mV} = 25 \times 10^{-3}~\text{V}$.

2. Determine the change in output current:

   The problem states that the collector current changes by $2~\text{mA}$. This is the change in current flowing through the load resistance.

   $\Delta I_C = 2~\text{mA} = 2 \times 10^{-3}~\text{A}$.

3. Identify the load resistance:

   The load resistance in the collector circuit is given as $5~\text{k}\Omega$.

   $R_L = 5~\text{k}\Omega = 5 \times 10^3~\Omega$.

4. Calculate the change in output voltage: …

The concept first: conservation is all you need

Every decay is bookkeeping on two conserved quantities — mass number $A$ (nucleons) and charge/atomic number $Z$ (protons):

• $\alpha$-decay ejects a $\ce{^{4}_{2}He}$ nucleus: $\;A \to A-4,\quad Z \to Z-2$.
• $\beta^{-}$-decay turns a neutron into a proton and ejects an electron: $\;A \to A$ (unchanged), $\;Z \to Z+1$.

And the quantity actually asked for, the neutron number, is never listed on the symbol — you compute it as

That last line is where most marks are lost: students correctly find R's $A$ and $Z$ and then quote $A$ instead of $A - Z$.

Step-by-step

Step 1 — First $\alpha$-decay ($X \to P$).

Step 2 — $\beta^{-}$-decay ($P \to Q$). Mass number frozen, $Z$ climbs by one:

Step 3 — Second $\alpha$-decay ($Q \to R$).

$^{A-4}_{Z-1}Q \;\longrightarrow\; ^{A-8}_{Z-3}R + ^{4}_{2}\text{He}$ …

The density of a nucleus is a fundamental property that reveals much about the strong nuclear force and the packing of nucleons (protons and neutrons) within the nucleus. A key insight in nuclear physics is that nuclear matter is incredibly dense and, remarkably, its density is nearly constant across all nuclei, irrespective of their size or mass number.

This constancy arises from two main observations:

1. Nuclear Mass: The mass of a nucleus is approximately proportional to its mass number ($A$), which is the total number of protons and neutrons. Each nucleon has roughly the same mass, approximately 1 atomic mass unit (amu). So, the total mass $M \approx A \times (1 \text{ amu})$.
2. Nuclear Volume: Experimental evidence shows that the radius ($R$) of a nucleus is approximately proportional to the cube root of its mass number ($A^{1/3}$). Since the nucleus is roughly spherical, its volume ($V$) is proportional to $R^3$, which means $V \propto (A^{1/3})^3 \propto A$.

When we calculate density ($\rho = M/V$), the proportionality to $A$ in both the numerator (mass) and the denominator (volume) cancels out. This leaves us with a density that is independent of $A$, meaning all nuclei have roughly the same density.

Let's calculate the approximate density for the given nucleus $\ce{^{60}_{30}X}$.

1. Identify the mass number and given constants:

   The nucleus is $\ce{^{60}_{30}X}$. The superscript $60$ is the mass number ($A$).

   So, $A = 60$.

   We are given:

   Mass of 1 amu $= 1.6 \times 10^{-27}$ kg

   $R_0 = 1.2 \times 10^{-15}$ m

2. Calculate the approximate mass of the nucleus:

   The mass of a nucleus is approximately the mass number multiplied by the mass of one nucleon (which is approximately 1 amu).

   $M = A \times (1 \text{ amu})$

   $M = 60 \times (1.6 \times 10^{-27} \text{ kg})$

   $M = 96 \times 10^{-27} \text{ kg}$

3. Calculate the volume of the nucleus:

   First, find the radius of the nucleus using the formula $R = R_0 A^{1/3}$.

   $R = (1.2 \times 10^{-15} \text{ m}) \times (60)^{1/3}$

   Since $60^{1/3} \approx 3.91$,

   $R \approx (1.2 \times 10^{-15} \text{ m}) \times 3.91$

   $R \approx 4.692 \times 10^{-15} \text{ m}$

   Now, calculate the volume of the nucleus, assuming it is a sphere:

   $V = \frac{4}{3}\pi R^3$

   $V = \frac{4}{3}\pi (4.692 \times 10^{-15} \text{ m})^3$

   $V \approx \frac{4}{3} \times 3.14159 \times (4.692)^3 \times (10^{-15})^3 \text{ m}^3$

   $V \approx \frac{4}{3} \times 3.14159 \times 103.1 \times 10^{-45} \text{ m}^3$

   $V \approx 431.6 \times 10^{-45} \text{ m}^3$

   $V \approx 4.316 \times 10^{-43} \text{ m}^3$

   Tip

   A more elegant way to calculate the volume is to substitute $R = R_0 A^{1/3}$ directly into the volume formula:

   $V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$.

   This shows explicitly how the volume is proportional to $A$.

   $V = \frac{4}{3}\pi (1.2 \times 10^{-15} \text{ m})^3 \times 60$

   $V = \frac{4}{3}\pi (1.728 \times 10^{-45} \text{ m}^3) \times 60$ …

The dynamic resistance $r_d$ of a p-n junction is defined as the small-signal resistance at a given operating point: it is the ratio of a small change in voltage to the resulting change in current, i.e. $r_d = \frac{dV}{dI}$. Physically, it tells you how much the voltage must shift to produce a unit current change around that bias — the steeper the I–V curve, the smaller the dynamic resistance.

For the given diode equation $I = I_0 (e^{V/V_T} - 1)$, the exponential term dominates once $V$ is more than a few $V_T$ above zero, so $I \approx I_0 e^{V/V_T}$. The dynamic resistance then follows directly from differentiation.

1. Differentiate the current expression with respect to $V$:

Since $I \gg I_0$ for any forward bias of interest, $\frac{dI}{dV} \approx \frac{I}{V_T}$.

2. The dynamic resistance is the reciprocal:

This is the key result: dynamic resistance is inversely proportional to the current.

3. Now evaluate at the two given currents:

4. The ratio is: $$$$

Concept and Intuition

In a p-n junction, a depletion region forms at the interface between the p-type and n-type semiconductors. This region is depleted of mobile charge carriers (electrons and holes) and contains immobile positive donor ions on the n-side and negative acceptor ions on the p-side. These fixed charges create an electric field that points from the n-side (positive ions) to the p-side (negative ions).

This electric field establishes a potential barrier across the depletion region. For an electron, which is negatively charged, moving from the n-side to the p-side means moving against the direction of the electric field. To do this, the electron must gain potential energy, or equivalently, it must do work against the electric field. This work done against the electric field is supplied by the electron's kinetic energy.

Therefore, the minimum kinetic energy an electron must possess to successfully cross the depletion region from the n-side to the p-side is exactly equal to the potential energy barrier it needs to overcome. This potential energy barrier is the work done by the electric field on the electron as it traverses the entire width of the depletion region.

Step-by-step Solution

1. Identify the given quantities:

   • Electric field strength in the depletion region, $E = 5 \times 10^5 \text{ V/m}$.
   • Minimum kinetic energy of a conduction electron to diffuse from n-side to p-side, $KE_{min} = 3.2 \times 10^{-20} \text{ J}$.
   • The charge of an electron, $q = e = 1.6 \times 10^{-19} \text{ C}$.

2. Relate kinetic energy to potential energy barrier:

   For an electron to move from the n-side to the p-side, it must overcome the potential energy barrier created by the electric field in the depletion region. The minimum kinetic energy required for this process is equal to the potential energy barrier.

   Let $\Delta U$ be the potential energy barrier. Then, $KE_{min} = \Delta U$.

3. Express potential energy barrier in terms of electric field and depletion width:

   The potential energy barrier $\Delta U$ for a charge $q$ moving across a potential difference $\Delta V$ is given by $\Delta U = q \Delta V$.

   In a region with a uniform electric field $E$ and width $d$, the potential difference $\Delta V$ across this region is given by $\Delta V = E d$. …

Concept and Intuition

A p-n junction is formed when a p-type semiconductor is brought into intimate contact with an n-type semiconductor. At the moment of formation, there's a significant difference in the concentration of charge carriers across the junction:

• The n-side has a high concentration of free electrons (majority carriers) and a low concentration of holes (minority carriers).
• The p-side has a high concentration of holes (majority carriers) and a low concentration of free electrons (minority carriers).

This concentration gradient drives a process called diffusion.

1. Electron Diffusion: Electrons from the n-side (high concentration) start moving across the junction to the p-side (low concentration).
2. Hole Diffusion: Similarly, holes from the p-side (high concentration) start moving across the junction to the n-side (low concentration).

As electrons diffuse from the n-side to the p-side, they leave behind positively charged, immobile donor ions in the n-region. As holes diffuse from the p-side to the n-side, they leave behind negatively charged, immobile acceptor ions in the p-region. This region, now containing immobile ions and depleted of free charge carriers, is called the depletion region.

The separation of these immobile charges creates an internal electric field across the depletion region, pointing from the n-side (positive ions) to the p-side (negative ions). This electric field opposes further diffusion and also causes a drift of minority carriers.

Let's analyze each statement based on these principles.

Step-by-Step Analysis

1. Analyze Statement (I): electrons are accumulated in the depletion region

   • When electrons diffuse from the n-side to the p-side, they leave behind positively charged, immobile donor ions in the n-side of the depletion region.
   • Similarly, holes diffusing from the p-side to the n-side leave behind negatively charged, immobile acceptor ions in the p-side of the depletion region.
   • The depletion region is defined by the presence of these immobile ions and the absence of free charge carriers (electrons and holes). It is depleted of free carriers, not accumulated with them.
   • Therefore, statement (I) is incorrect.

2. Analyze Statement (II): electron drift current is from $p$ – side to $n$ – side

   • The diffusion of majority carriers (electrons from n to p, holes from p to n) creates an electric field ($E$) across the depletion region. This electric field points from the positively charged n-side to the negatively charged p-side.
   • This electric field exerts a force on any free charge carriers.
   • Minority electrons present in the p-side are pushed by this electric field towards the n-side.
   • This movement of minority electrons constitutes an electron drift current from the p-side to the n-side.
   • Therefore, statement (II) is correct. …