Physics CSS Solved Papers

CSS Physics Paper-II 2022 Solved

Engr. Muhammad Yar Saqib

CSS Physics Paper-II 2022 Solved is a complete solved guide for CSS aspirants who need actual answers, not only short hints. This post solves the subjective section question by question with definitions, derivations, formulas, numerical substitutions, final answers, diagrams, tables and exam-ready explanations. It covers electric dipole field on the y-axis, vector expression of electric field, photoelectric effect, Einstein’s photoelectric equation, stopping potential, intrinsic and extrinsic semiconductors, magnetic materials, Landé g factor, electric field at a square corner due to other charges, parallel plate capacitor area, particle in a one-dimensional box, wave functions and probability densities, charged particle motion in magnetic field, atomic description of dielectrics, dielectric slab capacitor, antiparticles, beta radiation detection and proof from uncertainty principle that electrons do not pre-exist inside the nucleus.

Central Argument: A CSS Physics solved paper should not only give final answers. It should show the physical principle, the derivation path, the mathematical substitution, the final unit and the exam-writing logic. Therefore, this CSS Physics Paper-II 2022 Solved post gives the full route to each answer so students can reproduce the solution in the examination hall.

Scan Note: The uploaded draft contains OCR distortions such as , , , , , and $- radiation. In this corrected solution, these are interpreted as the standard physics symbols: charge q, half-separation a, electric field E, distance r, source charge Q, and beta-minus radiation β⁻. In Q5(a), the printed 4p is treated as OCR distortion of 4q.

CSS Physics Paper-II 2022 Solved Study Scope

This post covers CSS Physics Paper-II 2022 Solved in a complete, structured and WordPress-ready format. Each question includes the printed question, part-wise solution, formulas, mathematical working, final answer and examiner-friendly explanation.

Use this solved paper as a study document. First revise the formula sheet, then read Q2 to Q8 one by one. After reading, close the post and rewrite each solution from memory. CSS Physics rewards clear definitions, correct assumptions, dimensional accuracy, proper units and disciplined presentation.

Show Table of Contents
  1. Overview
  2. Study Scope
  3. Important Formula Sheet
  4. Question Map
  5. Question 2: Electric Dipole Field, Point-Charge Field and Definitions
  6. Question 3: Photoelectric Effect and Stopping Potential
  7. Question 4: Semiconductors, Magnetic Materials and Landé g Factor
  8. Question 5: Square Charges, Capacitor Area and Definitions
  9. Question 6: Particle in a One-Dimensional Box
  10. Question 7: Charged Particle in Magnetic Field and Dielectric Slab Capacitor
  11. Question 8: Antiparticles, Beta Radiation Detection and Electron Outside Nucleus
  12. Revision Plan
  13. Internal and External Resources
  14. FAQs

Important Formula Sheet for CSS Physics Paper-II 2022 Solved

Electrostatics

E = F/q0

E = (1/4πε0)(q/r²) r̂

E = (1/4πε0)(q r/r³)

p = 2qa

C = ε0A/d

Modern Physics

Eγ = hf = hc/λ

Kmax = hf - φ

eVs = Kmax

λ = h/p

ΔxΔp ≥ ℏ/2

Quantum Mechanics

-(ℏ²/2m)d²ψ/dx² = Eψ

ψn = √(2/L) sin(nπx/L)

En = n²h²/(8mL²)

|ψn|² = (2/L)sin²(nπx/L)

Magnetism and Dielectrics

F = q(v×B)

r = mv/(|q|B)

ω = |q|B/m

C' = ε0A/(d-t+t/K)

C' = C0d/(d-t+t/K)

Question Map of CSS Physics Paper-II 2022 Solved

Question Main Area What Is Fully Solved
Q2 Electrostatics Electric field of dipole on y-axis, point-charge electric field vector and definitions of electric field and dipole.
Q3 Modern physics Photoelectric effect, Einstein’s equation, failure of wave theory and stopping potential numerical.
Q4 Solid-state and magnetism Intrinsic/extrinsic semiconductors, diamagnetic/paramagnetic/ferromagnetic materials and Landé g factor.
Q5 Electrostatics Electric field at square corner due to other charges, force on q, 1 F capacitor plate area and definitions.
Q6 Quantum mechanics Schrödinger equation for particle in box, eigenfunctions, energy levels and probability density graphs.
Q7 Magnetism and dielectrics Charged particle motion in magnetic field, atomic description of dielectrics and dielectric slab capacitance.
Q8 Nuclear and particle physics Particle-antiparticle pairs, beta-minus radiation detection and proof that electron does not exist inside nucleus.

Question 2: Electric Dipole Field, Point-Charge Field and Definitions

Full Question

Q.2. (a) An electric dipole, comprising a positive charge +q and a negative charge -q, is placed on the x-axis. Each charge is at the same distance from the origin. The total separation between the charges is 2a. Calculate the electric field E due to these charges along the y-axis at point P, which is at a distance y from the origin. Assume y >> a and ε0 = 8.85×10⁻¹² C²/Nm².

Q.2. (b) Write down a mathematical expression to evaluate electric field E at a distance r from the source charge Q in vector form. Sketch the graph of E as a function of r.

Q.2. (c) Define electric field and a dipole.

Q2(a): Electric Field of a Dipole on the y-axis

Let the dipole lie on the x-axis with charges +q and -q separated by distance 2a. The point P lies on the y-axis at distance y from the origin.

Assume:

+q is at x = +a
-q is at x = -a
P is at (0, y)

The distance of each charge from point P is:

R = √(a² + y²)

The electric field due to +q at P is directed away from +q. Its vector is:

E+ = (1/4πε0) q(-a i + y j)/(a²+y²)^(3/2)

The electric field due to -q at P is directed toward -q. Its vector is:

E- = (1/4πε0)(-q)(a i + y j)/(a²+y²)^(3/2)
E- = (1/4πε0) q(-a i – y j)/(a²+y²)^(3/2)

Now add both fields:

E = E+ + E-
E = (1/4πε0) q[(-a i + y j) + (-a i – y j)]/(a²+y²)^(3/2)
E = (1/4πε0) q[-2a i]/(a²+y²)^(3/2)

Therefore:

E = -(1/4πε0) (2qa)/(a²+y²)^(3/2) i

The dipole moment magnitude is:

p = 2qa

So:

E = -(1/4πε0) p/(a²+y²)^(3/2) i

For y >> a:

(a²+y²)^(3/2) ≈ y³

Hence:

E ≈ -(1/4πε0)(p/y³)i
Final Answer for Q2(a): On the equatorial line of the dipole, E = -(1/4πε0)p/y³ for y >> a. The direction is opposite to the dipole moment if the dipole moment points from -q to +q.

Q2(b): Electric Field of a Point Charge in Vector Form

The electric field at a distance r from a point charge Q is:

E = (1/4πε0)(Q/r²) r̂

In full vector form:

E(r) = (1/4πε0)(Q r/r³)

where r is the position vector from the source charge to the field point.

Graph of Electric Field with Distance

The magnitude is:

E ∝ 1/r²

Therefore, electric field decreases rapidly as distance increases.

E
^
|\
| \
| \
| \
| \
| \______
| \______
+————————–> r

E is very large near the charge and decreases as 1/r².

Final Answer for Q2(b): E=(1/4πε0)(Q/r²)r̂. The graph of E against r is an inverse-square curve.

Q2(c): Definitions

Electric Field

Electric field at a point is the force experienced by a unit positive test charge placed at that point.

E = F/q0

Its SI unit is:

N/C or V/m

Electric Dipole

An electric dipole is a system of two equal and opposite charges separated by a small distance.

If charges +q and -q are separated by distance 2a, dipole moment is:

p = q(2a) = 2qa

The direction of dipole moment is from negative charge to positive charge.

Final Answer for Q2(c): Electric field is force per unit positive test charge. Electric dipole is a pair of equal and opposite charges separated by a small distance.

Question 3: Photoelectric Effect and Stopping Potential

Full Question

Q.3. (a) Discuss photoelectric effect and establish Einstein’s equation for the photoelectric effect.

Q.3. (b) Describe the inadequacy of the wave theory of light to explain the effect.

Q.3. (c) A photon of energy 12 eV falls on a certain metal plate whose work function is 4.15 eV. Find the stopping potential. The mass and charge of electron are 9.11×10⁻³¹ kg and 1.6×10⁻¹⁹ C respectively, and Planck’s constant is 6.64×10⁻³⁴ J s.

Q3(a): Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

Important observations are:

  1. Emission occurs only if incident light has frequency greater than a threshold frequency.
  2. Photoemission is almost instantaneous.
  3. Maximum kinetic energy of photoelectrons depends on frequency, not intensity.
  4. Photoelectric current depends on intensity if frequency is above threshold.

Einstein’s Photoelectric Equation

Einstein explained the photoelectric effect by assuming that light consists of photons. Each photon has energy:

E = hf

When one photon strikes one electron, its energy is used in two parts:

  1. To overcome the work function φ of the metal.
  2. To give maximum kinetic energy Kmax to the emitted electron.

Therefore:

hf = φ + Kmax

So:

Kmax = hf – φ

If stopping potential is Vs, then:

Kmax = eVs

Thus:

eVs = hf – φ
Final Answer for Q3(a): Einstein’s photoelectric equation is hf = φ + Kmax, or Kmax = hf - φ.

Q3(b): Inadequacy of Wave Theory

The classical wave theory of light could not explain several facts of the photoelectric effect.

Observation Why Wave Theory Fails Photon Explanation
Threshold frequency exists. Wave theory predicts enough intensity should eventually eject electrons at any frequency. Each photon must have energy hf ≥ φ.
Emission is instantaneous. Wave theory expects delay while electron absorbs energy gradually. Electron absorbs energy from one photon at once.
Kinetic energy depends on frequency. Wave theory predicts kinetic energy should depend mainly on intensity. Kmax = hf - φ.
Intensity affects current, not maximum energy. Wave theory cannot separate current and kinetic energy properly. Intensity means more photons, so more electrons are emitted.
Final Answer for Q3(b): Wave theory fails because it cannot explain threshold frequency, instantaneous emission and frequency dependence of photoelectron kinetic energy.

Q3(c): Stopping Potential Numerical

Given:

Photon energy = 12 eV
Work function, φ = 4.15 eV

Einstein’s equation gives:

Kmax = Eγ – φ

Substitute values:

Kmax = 12 – 4.15
Kmax = 7.85 eV

Stopping potential is related to kinetic energy by:

eVs = Kmax

When kinetic energy is in electron-volts, stopping potential in volts has the same numerical value:

Vs = 7.85 V
Final Answer for Q3(c): The stopping potential is 7.85 V.

Question 4: Semiconductors, Magnetic Materials and Landé g Factor

Full Question

Q.4. (a) Discuss intrinsic and extrinsic semiconductors.

Q.4. (b) Describe the properties of diamagnetic, paramagnetic and ferromagnetic materials.

Q.4. (c) Briefly discuss the Landé g factor.

Q4(a): Intrinsic and Extrinsic Semiconductors

A semiconductor is a material whose conductivity lies between that of a conductor and an insulator. Silicon and germanium are common examples.

Intrinsic Semiconductor

An intrinsic semiconductor is a pure semiconductor with no intentional impurity added. In an intrinsic semiconductor, electron-hole pairs are generated thermally.

Important properties are:

  1. It is chemically pure.
  2. Number of electrons equals number of holes.
  3. Conductivity is low at low temperature.
  4. Conductivity increases with temperature.
n = p = ni

Extrinsic Semiconductor

An extrinsic semiconductor is formed by adding a small controlled amount of impurity to an intrinsic semiconductor. This process is called doping.

n-Type Semiconductor

n-type semiconductor is made by doping silicon or germanium with pentavalent impurity atoms such as phosphorus, arsenic or antimony. These atoms donate extra electrons.

n-Type: Majority carriers are electrons; minority carriers are holes.

p-Type Semiconductor

p-type semiconductor is made by doping with trivalent impurity atoms such as boron, aluminium, gallium or indium. These atoms create holes.

p-Type: Majority carriers are holes; minority carriers are electrons.
Feature Intrinsic Semiconductor Extrinsic Semiconductor
Purity Pure semiconductor. Doped semiconductor.
Carrier source Thermal generation. Dopant atoms.
Carrier concentration n=p Either electrons or holes dominate.
Conductivity Low. Higher than intrinsic.
Types No n/p type. n-type and p-type.
Final Answer for Q4(a): Intrinsic semiconductors are pure with equal electrons and holes. Extrinsic semiconductors are doped and may be n-type or p-type depending on the impurity.

Q4(b): Diamagnetic, Paramagnetic and Ferromagnetic Materials

Diamagnetic Materials

Diamagnetic materials are weakly repelled by a magnetic field. They develop induced magnetic moments opposite to the applied field.

  • Magnetic susceptibility is small and negative.
  • Relative permeability is slightly less than 1.
  • They do not retain magnetism after field removal.
  • Examples: bismuth, copper, silver, water.

Paramagnetic Materials

Paramagnetic materials are weakly attracted by a magnetic field because they have unpaired electrons.

  • Magnetic susceptibility is small and positive.
  • Relative permeability is slightly greater than 1.
  • Magnetization disappears when field is removed.
  • Examples: aluminium, platinum, oxygen.

Ferromagnetic Materials

Ferromagnetic materials are strongly attracted by a magnetic field and can retain magnetization after the external field is removed.

  • Magnetic susceptibility is large and positive.
  • They contain magnetic domains.
  • They show hysteresis.
  • They can be permanently magnetized.
  • Examples: iron, cobalt, nickel.
Property Diamagnetic Paramagnetic Ferromagnetic
Response to field Weakly repelled Weakly attracted Strongly attracted
Susceptibility Negative Small positive Large positive
Magnetic domains No domain alignment No permanent domain alignment Strong domain alignment
Retains magnetism No No Yes
Examples Cu, Bi, water Al, Pt, O₂ Fe, Co, Ni
Final Answer for Q4(b): Diamagnetic materials weakly oppose a magnetic field, paramagnetic materials weakly align with it, and ferromagnetic materials strongly align and can retain magnetization.

Q4(c): Landé g Factor

The Landé g factor is a dimensionless factor that relates the total angular momentum of an atom to its magnetic moment. It is important in atomic spectroscopy, especially in the Zeeman effect.

The Landé g factor for an atom in LS coupling is:

gJ = 1 + [J(J+1) + S(S+1) – L(L+1)]/[2J(J+1)]

where:

  • L is total orbital angular momentum quantum number.
  • S is total spin angular momentum quantum number.
  • J is total angular momentum quantum number.

Use in Zeeman Effect

The energy shift of an atomic level in a magnetic field is:

ΔE = μB gJ mJ B

where μB is Bohr magneton, mJ is magnetic quantum number and B is magnetic field.

Final Answer for Q4(c): Landé g factor measures the relationship between total angular momentum and magnetic moment of an atom and explains splitting of spectral lines in a magnetic field.

Question 5: Square Charges, Capacitor Area and Definitions

Full Question

Q.5. (a) Four charged particles of charge q, 2q, 3q and 4q are at the corners of a square of side a arranged in counter-clockwise direction. Determine: (i) the electric field at the location of charge q, and (ii) the total electric force exerted on q.

Q.5. (b) A parallel plate capacitor has a plate separation of 1 mm. Calculate the surface area of each plate of the capacitor to obtain a capacitance of 1 F. Is it possible to produce such a capacitor in the lab? Comment. Given ε0 = 8.85×10⁻¹² C²/Nm².

Q.5. (c) Define: (i) capacitance, (ii) unit of capacitance, and (iii) surface charge density.

Q5(a): Electric Field at the Location of Charge q

Assumption: The printed text appears as 4p, but in context it should be 4q. The solution assumes four positive charges q, 2q, 3q and 4q placed counter-clockwise at the corners of a square.

Let the square have side a. Place charge q at the origin:

q at (0,0)
2q at (a,0)
3q at (a,a)
4q at (0,a)

We calculate electric field at the position of q due to the other three charges only.

Let:

k = 1/(4πε0)

Field Due to 2q

Charge 2q is at distance a along +x direction from the origin. At the origin, its field points in the negative x-direction.

E2q = – k(2q/a²) i

Field Due to 4q

Charge 4q is at distance a along +y direction from the origin. At the origin, its field points in the negative y-direction.

E4q = – k(4q/a²) j

Field Due to 3q

Charge 3q is at the opposite corner (a,a). Its distance from the origin is:

r = a√2

Magnitude of field is:

E3q = k(3q)/(a√2)²
E3q = k(3q)/(2a²)

Direction from 3q to the origin is along:

(-i – j)/√2

Therefore:

E3q = [k(3q)/(2a²)] [(-i-j)/√2]
E3q = – k(3q)/(2√2 a²)i – k(3q)/(2√2 a²)j

Total Electric Field at q

Add all components:

E = E2q + E4q + E3q
E = – kq/a² [2 + 3/(2√2)] i – kq/a² [4 + 3/(2√2)] j
Electric Field at q: E = -(kq/a²)[2+3/(2√2)]i -(kq/a²)[4+3/(2√2)]j.

Total Force on q

Force on charge q is:

F = qE

Therefore:

F = – kq²/a² [2 + 3/(2√2)] i – kq²/a² [4 + 3/(2√2)] j
Final Answer for Q5(a): F = -(kq²/a²)[2+3/(2√2)]i -(kq²/a²)[4+3/(2√2)]j.

Q5(b): Surface Area for 1 F Parallel Plate Capacitor

For a parallel plate capacitor:

C = ε0A/d

Given:

C = 1 F
d = 1 mm = 1×10⁻³ m
ε0 = 8.85×10⁻¹² F/m

Rearrange for area:

A = Cd/ε0

Substitute values:

A = (1)(1×10⁻³)/(8.85×10⁻¹²)
A = 1.13×10⁸ m²
Final Answer for Q5(b): Required plate area is approximately 1.13×10⁸ m² for each plate.

Is it Possible in the Lab?

No, it is not practical for an ordinary lab capacitor. The required area is extremely large. This shows why a simple air-filled parallel plate capacitor cannot easily provide capacitance as large as 1 F. Practical capacitors use very thin dielectrics, rolled plates, porous electrodes, electrolytic structures or supercapacitor designs.

Q5(c): Definitions

Capacitance

Capacitance is the ability of a conductor or capacitor to store charge per unit potential difference.

C = Q/V

Unit of Capacitance

The SI unit of capacitance is farad.

1 farad = 1 coulomb/volt

Surface Charge Density

Surface charge density is charge per unit area on a surface.

σ = Q/A

Its SI unit is:

C/m²
Final Answer for Q5(c): Capacitance is C=Q/V, its unit is farad, and surface charge density is σ=Q/A.

Question 6: Particle in a One-Dimensional Box

Full Question

Q.6. (a) Set up the Schrödinger wave equation for a particle of mass m confined in a one-dimensional box with perfectly rigid walls at x=0 and x=L. Solve the differential equation to find expressions for energy and eigen wave functions of the particle.

Q.6. (b) Sketch the graphs for the first three eigen wave functions ψ1, ψ2 and ψ3.

Q.6. (c) Plot the graphs for the probability densities corresponding to ψ1, ψ2 and ψ3.

Q6(a): Schrödinger Equation for Particle in a Box

A particle in a one-dimensional infinite potential well is confined between x=0 and x=L. The potential energy is:

V(x) = 0, 0 < x < L
V(x) = ∞, x ≤ 0 and x ≥ L

Inside the box, the time-independent Schrödinger equation is:

-(ℏ²/2m)d²ψ/dx² = Eψ

Rearrange:

d²ψ/dx² + (2mE/ℏ²)ψ = 0

Let:

k² = 2mE/ℏ²

Then:

d²ψ/dx² + k²ψ = 0

The general solution is:

ψ(x) = A sin(kx) + B cos(kx)

Boundary Conditions

Since the walls are infinitely rigid, wave function must be zero at the walls:

ψ(0) = 0

Apply x=0:

ψ(0) = A sin0 + B cos0 = B

So:

B = 0

Thus:

ψ(x) = A sin(kx)

Now apply ψ(L)=0:

ψ(L) = A sin(kL) = 0

For a non-zero wave function:

sin(kL) = 0

Therefore:

kL = nπ
k = nπ/L, n = 1,2,3,…

Energy Eigenvalues

Since:

k² = 2mE/ℏ²

we get:

E = ℏ²k²/(2m)

Substitute k=nπ/L:

En = ℏ²n²π²/(2mL²)

Using ℏ=h/2π:

En = n²h²/(8mL²)

Normalized Eigenfunctions

The normalized wave functions are:

ψn(x) = √(2/L) sin(nπx/L)
Final Answer for Q6(a): ψn=√(2/L)sin(nπx/L) and En=n²h²/(8mL²), where n=1,2,3,....

Q6(b): First Three Eigen Wave Functions

ψ1 = √(2/L) sin(πx/L)
ψ2 = √(2/L) sin(2πx/L)
ψ3 = √(2/L) sin(3πx/L)
Wave functions in a one-dimensional box

ψ1:
0 /\
| / \
|____/____\____ x
0 L

ψ2:
0 /\ /\
| / \ / \
|___/____\__/____\___ x
0 L/2 L

ψ3:
0 /\ /\ /\
| / \ / \ / \
|__/____\/____\/____\__ x
0 L/3 2L/3 L

The first wave function has one half-wave, the second has two half-waves, and the third has three half-waves.

Q6(c): Probability Density Graphs

Probability density is:

|ψn|² = (2/L) sin²(nπx/L)

Therefore:

|ψ1|² = (2/L) sin²(πx/L)
|ψ2|² = (2/L) sin²(2πx/L)
|ψ3|² = (2/L) sin²(3πx/L)
Probability densities

|ψ1|²:
0 /\
| / \
|____/____\____ x
0 L
One probability maximum.

|ψ2|²:
0 /\ /\
| / \ / \
|__/____\/____\__ x
0 L/2 L
Two probability maxima.

|ψ3|²:
0 /\ /\ /\
| / \ / \ / \
|_/____V____V____\_ x
0 L/3 2L/3 L
Three probability maxima.

Final Answer for Q6(b-c): The first three wave functions have 1, 2 and 3 half-waves respectively. Their probability densities are sine-square curves with 1, 2 and 3 peaks respectively.

Question 7: Charged Particle in Magnetic Field and Dielectric Slab Capacitor

Full Question

Q.7. (a) Discuss the motion of a charged particle of mass m, charge q and velocity v in a magnetic field B directed into the plane of the paper.

Q.7. (b) Discuss atomic description of dielectrics.

Q.7. (c) Let d be the separation between the parallel plates of a capacitor of capacitance C0 in the absence of dielectric material. A slab of material of dielectric constant K and thickness t < d is placed between the plates. Calculate the capacitance in the presence of the dielectric material.

Q7(a): Motion of a Charged Particle in Magnetic Field

The magnetic force on a moving charged particle is:

F = q(v × B)

If velocity v is perpendicular to magnetic field B, the magnitude of force is:

F = |q|vB

This magnetic force is always perpendicular to velocity. Therefore, it does not change the speed of the particle. It only changes the direction of motion and acts as centripetal force.

For circular motion:

|q|vB = mv²/r

Cancel one v:

|q|B = mv/r

So radius is:

r = mv/(|q|B)

Angular Frequency and Period

Angular speed is:

ω = v/r

Substitute r=mv/(|q|B):

ω = |q|B/m

Time period is:

T = 2π/ω
T = 2πm/(|q|B)

Direction of Motion

If the magnetic field is directed into the plane of the paper, the direction of force is found by the right-hand rule for a positive charge. A negative charge moves in the opposite sense.

Final Answer for Q7(a): A charged particle moving perpendicular to a uniform magnetic field follows circular motion with r=mv/(|q|B), ω=|q|B/m, and T=2πm/(|q|B).

Q7(b): Atomic Description of Dielectrics

A dielectric is an insulating material that becomes polarized when placed in an electric field. It does not allow free flow of charge like a conductor, but its internal charges shift slightly under an applied field.

Atomic Polarization

In an atom, the positively charged nucleus and negatively charged electron cloud have the same centre in the absence of an external electric field. When an electric field is applied, the electron cloud shifts slightly opposite to the field and the nucleus shifts slightly along the field. This creates an induced dipole.

Polar Molecules

Some molecules already have permanent dipole moments. In the absence of field, these dipoles are randomly oriented. In an electric field, they tend to align with the field, producing orientational polarization.

Types of Polarization

Type Explanation
Electronic polarization Displacement of electron cloud relative to nucleus.
Ionic polarization Relative displacement of positive and negative ions.
Orientational polarization Alignment of permanent molecular dipoles.
Space-charge polarization Charge accumulation at boundaries or defects.

Effect on Capacitor

When a dielectric is placed inside a capacitor, it reduces the effective electric field for the same free charge and increases capacitance.

C = K C0

This expression applies when the space between plates is completely filled with dielectric.

Final Answer for Q7(b): Dielectrics polarize because charges inside atoms or molecules shift slightly or dipoles align in an external electric field.

Q7(c): Capacitance with Dielectric Slab of Thickness t

Let plate separation be d, plate area be A, and original capacitance without dielectric be:

C0 = ε0A/d

A dielectric slab of thickness t and dielectric constant K is inserted between the plates. The remaining air gap is:

d – t

The system behaves like two capacitors in series:

  • Air region of thickness d-t
  • Dielectric region of thickness t

The effective separation becomes:

deff = (d – t) + t/K

Therefore, capacitance is:

C’ = ε0A / [(d-t) + t/K]

Using C0=ε0A/d, we can write:

ε0A = C0d

So:

C’ = C0d / [d – t + t/K]
Final Answer for Q7(c): C' = ε0A/(d-t+t/K), or C' = C0d/(d-t+t/K).

Special Case

If dielectric completely fills the space, then t=d:

C’ = C0d/[d/K] = KC0

This matches the standard result for a fully filled dielectric capacitor.

Question 8: Antiparticles, Beta Radiation Detection and Electron Outside Nucleus

Full Question

Q.8. (a) Discuss the properties of three subatomic particles and their corresponding antiparticles.

Q.8. (b) Explain in detail how β⁻ radiation can be detected.

Q.8. (c) How can we prove that an electron does not exist in the nucleus of an atom?

Q8(a): Subatomic Particles and Their Antiparticles

An antiparticle has the same mass as its corresponding particle but opposite charge and opposite relevant quantum numbers. When a particle meets its antiparticle, they may annihilate and convert mass into energy.

Particle Charge Antiparticle Antiparticle Charge Important Property
Electron e⁻ -e Positron e⁺ +e Same mass, opposite charge.
Proton p +e Antiproton -e Same mass, opposite charge and baryon number.
Neutron n 0 Antineutron 0 Same mass but opposite baryon number and magnetic moment.

Annihilation Example

Electron and positron can annihilate:

e⁻ + e⁺ → γ + γ

This process converts rest mass into electromagnetic radiation while conserving energy, momentum and charge.

Final Answer for Q8(a): Electron-positron, proton-antiproton and neutron-antineutron are particle-antiparticle pairs. Antiparticles have same mass but opposite charge or opposite quantum numbers.

Q8(b): Detection of Beta-Minus Radiation

Beta-minus radiation consists of high-speed electrons emitted from unstable nuclei. In beta-minus decay:

n → p + e⁻ + ν̄

Beta particles can be detected because they ionize matter and are deflected by electric and magnetic fields.

1. Geiger-Muller Counter

A Geiger-Muller tube contains gas at low pressure. When beta particles enter the tube, they ionize gas atoms. The resulting ions and electrons create an electrical pulse, which is counted by the circuit.

Detection Steps

  1. Beta particle enters the GM tube through a thin window.
  2. It ionizes gas molecules.
  3. Electrons accelerate toward the anode wire.
  4. An avalanche of ionization occurs.
  5. A voltage pulse is produced and counted.

2. Cloud Chamber

In a cloud chamber, beta particles ionize supersaturated vapour. Droplets form along the ionized path, making the track visible. Beta particles produce thin, irregular tracks because they are light and easily scattered.

3. Scintillation Counter

In a scintillation detector, beta radiation strikes a scintillating material and produces tiny flashes of light. These flashes are converted into electrical pulses by a photomultiplier tube.

4. Magnetic Deflection

Beta-minus particles are electrons, so they are deflected in magnetic and electric fields. Their deflection direction confirms their negative charge.

5. Absorption Method

Beta particles have moderate penetrating power. They can pass through paper but are absorbed by a few millimetres of aluminium. Absorption tests help distinguish beta radiation from alpha and gamma radiation.

Detector / Method How It Detects β⁻ Radiation
GM counter Counts ionization pulses.
Cloud chamber Shows visible ionization tracks.
Scintillation counter Converts light flashes into pulses.
Magnetic field Shows negative charge by deflection.
Absorber test Beta is stopped by thin aluminium.
Final Answer for Q8(b): Beta-minus radiation can be detected by GM counter, cloud chamber, scintillation detector, magnetic deflection and absorption tests.

Q8(c): Proof That Electron Does Not Exist Inside Nucleus

One strong argument comes from Heisenberg’s uncertainty principle. If an electron existed inside the nucleus, it would be confined to a very small region of size about:

Δx ≈ 10⁻¹⁵ m

Uncertainty principle states:

Δx Δp ≥ ℏ/2

So:

Δp ≥ ℏ/(2Δx)

Using ℏ = 1.055×10⁻³⁴ J s:

Δp ≥ (1.055×10⁻³⁴)/(2×10⁻¹⁵)
Δp ≥ 5.275×10⁻²⁰ kg m/s

The corresponding relativistic energy would be approximately:

E ≈ pc
E ≈ (5.275×10⁻²⁰)(3×10⁸)
E ≈ 1.58×10⁻¹¹ J

Convert to electron-volts:

E ≈ (1.58×10⁻¹¹)/(1.6×10⁻¹⁹) eV
E ≈ 9.9×10⁷ eV
E ≈ 99 MeV

This energy is far greater than typical beta-decay electron energies. Therefore, an electron cannot be sitting inside the nucleus before decay.

Modern Explanation of Beta Decay

In beta-minus decay, the electron is created during the decay process through weak interaction:

n → p + e⁻ + ν̄

The electron is not a pre-existing nuclear constituent. It is produced at the moment of beta decay.

Final Answer for Q8(c): If an electron were confined inside the nucleus, uncertainty principle would give it energy of the order of 100 MeV, much larger than observed beta energies. Therefore, electrons do not pre-exist inside nuclei; they are created during beta decay.

Revision Plan for CSS Physics Paper-II 2022 Solved

After reading this complete CSS Physics Paper-II 2022 Solved guide, revise it in three rounds. In the first round, learn definitions and formulas. In the second round, reproduce derivations without looking. In the third round, solve the numerical questions again with units.

Question Revision Task
Q2 Derive dipole field on y-axis, write point-charge field vector and draw inverse-square graph.
Q3 Write Einstein photoelectric equation, explain wave-theory failure and solve stopping potential.
Q4 Compare intrinsic/extrinsic semiconductors, magnetic materials and memorize Landé g factor use.
Q5 Resolve the square-charge vector field and recalculate capacitor plate area for 1 F.
Q6 Derive particle-in-box eigenfunctions and energy levels; sketch ψ and |ψ|².
Q7 Derive circular motion in magnetic field and dielectric slab capacitance formula.
Q8 Prepare antiparticle table, beta detection methods and uncertainty proof for electron outside nucleus.

Related Resources for CSS Physics Paper-II 2022 Solved

Exam Note: For CSS Physics Paper-II, many questions mix theory, definitions and mathematical derivation. Begin with the physical principle, write the formula, derive cleanly, show units in numericals and end with a one-line physical interpretation.

FAQs About CSS Physics Paper-II 2022 Solved

What does CSS Physics Paper-II 2022 Solved include?

CSS Physics Paper-II 2022 Solved includes complete solved answers for Q2 to Q8 with definitions, derivations, numerical calculations, formulas, final answers, diagrams and exam-ready notes.

What is the electric field of a dipole on the y-axis?

For a dipole of moment p=2qa on the x-axis, the field on the y-axis for y>>a is E=-(1/4πε0)(p/y³), opposite to the dipole moment.

What is the stopping potential in Question 3?

The photon energy is 12 eV and the work function is 4.15 eV. Therefore, Kmax=7.85 eV and stopping potential is 7.85 V.

What area is required for a 1 F parallel plate capacitor with 1 mm separation?

The required area is A=Cd/ε0=1.13×10⁸ m², which is not practical for an ordinary lab capacitor.

What are the energy levels of a particle in a one-dimensional box?

The energy levels are En=n²h²/(8mL²), where n=1,2,3,....

What is the radius of a charged particle moving in a magnetic field?

For motion perpendicular to a uniform magnetic field, radius is r=mv/(|q|B).

What is capacitance with a dielectric slab of thickness t?

If a dielectric slab of thickness t and dielectric constant K is inserted in a capacitor of plate separation d, capacitance becomes C'=ε0A/(d-t+t/K).

How is beta-minus radiation detected?

Beta-minus radiation can be detected by GM counter, cloud chamber, scintillation counter, magnetic deflection and absorption in thin aluminium.

Why does an electron not exist inside the nucleus?

If an electron were confined inside the nucleus, uncertainty principle would give it energy around 100 MeV, far greater than observed beta-decay electron energies. Therefore, electrons are created during beta decay rather than pre-existing inside the nucleus.

Can I paste this HTML into WordPress?

Yes. The post avoids an H1 heading so your WordPress post title can remain the only H1. The internal structure begins with H2 and continues with H3/H4 headings.

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