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
- Overview
- Study Scope
- Important Formula Sheet
- Question Map
- Question 2: Electric Dipole Field, Point-Charge Field and Definitions
- Question 3: Photoelectric Effect and Stopping Potential
- Question 4: Semiconductors, Magnetic Materials and Landé g Factor
- Question 5: Square Charges, Capacitor Area and Definitions
- Question 6: Particle in a One-Dimensional Box
- Question 7: Charged Particle in Magnetic Field and Dielectric Slab Capacitor
- Question 8: Antiparticles, Beta Radiation Detection and Electron Outside Nucleus
- Revision Plan
- Internal and External Resources
- 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
P is at (0, y)
The distance of each charge from point P is:
The electric field due to +q at P is directed away from +q. Its vector is:
The electric field due to -q at P is directed toward -q. Its vector is:
Now add both fields:
Therefore:
The dipole moment magnitude is:
So:
For y >> a:
Hence:
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:
In full vector form:
where r is the position vector from the source charge to the field point.
Graph of Electric Field with Distance
The magnitude is:
Therefore, electric field decreases rapidly as distance increases.
^
|\
| \
| \
| \
| \
| \______
| \______
+————————–> r
E is very large near the charge and decreases as 1/r².
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.
Its SI unit is:
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:
The direction of dipole moment is from negative charge to positive charge.
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:
- Emission occurs only if incident light has frequency greater than a threshold frequency.
- Photoemission is almost instantaneous.
- Maximum kinetic energy of photoelectrons depends on frequency, not intensity.
- 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:
When one photon strikes one electron, its energy is used in two parts:
- To overcome the work function
φof the metal. - To give maximum kinetic energy
Kmaxto the emitted electron.
Therefore:
So:
If stopping potential is Vs, then:
Thus:
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. |
Q3(c): Stopping Potential Numerical
Given:
Work function, φ = 4.15 eV
Einstein’s equation gives:
Substitute values:
Stopping potential is related to kinetic energy by:
When kinetic energy is in electron-volts, stopping potential in volts has the same numerical value:
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:
- It is chemically pure.
- Number of electrons equals number of holes.
- Conductivity is low at low temperature.
- Conductivity increases with temperature.
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.
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.
| 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. |
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 |
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:
where:
Lis total orbital angular momentum quantum number.Sis total spin angular momentum quantum number.Jis total angular momentum quantum number.
Use in Zeeman Effect
The energy shift of an atomic level in a magnetic field is:
where μB is Bohr magneton, mJ is magnetic quantum number and B is 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:
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:
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.
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.
Field Due to 3q
Charge 3q is at the opposite corner (a,a). Its distance from the origin is:
Magnitude of field is:
Direction from 3q to the origin is along:
Therefore:
Total Electric Field at q
Add all components:
E = -(kq/a²)[2+3/(2√2)]i -(kq/a²)[4+3/(2√2)]j.Total Force on q
Force on charge q is:
Therefore:
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:
Given:
d = 1 mm = 1×10⁻³ m
ε0 = 8.85×10⁻¹² F/m
Rearrange for area:
Substitute values:
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.
Unit of Capacitance
The SI unit of capacitance is farad.
Surface Charge Density
Surface charge density is charge per unit area on a surface.
Its SI unit 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) = ∞, x ≤ 0 and x ≥ L
Inside the box, the time-independent Schrödinger equation is:
Rearrange:
Let:
Then:
The general solution is:
Boundary Conditions
Since the walls are infinitely rigid, wave function must be zero at the walls:
Apply x=0:
So:
Thus:
Now apply ψ(L)=0:
For a non-zero wave function:
Therefore:
Energy Eigenvalues
Since:
we get:
Substitute k=nπ/L:
Using ℏ=h/2π:
Normalized Eigenfunctions
The normalized wave functions are:
ψn=√(2/L)sin(nπx/L) and En=n²h²/(8mL²), where n=1,2,3,....Q6(b): First Three Eigen Wave Functions
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:
Therefore:
|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.
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:
If velocity v is perpendicular to magnetic field B, the magnitude of force is:
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:
Cancel one v:
So radius is:
Angular Frequency and Period
Angular speed is:
Substitute r=mv/(|q|B):
Time period is:
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.
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.
This expression applies when the space between plates is completely filled with dielectric.
Q7(c): Capacitance with Dielectric Slab of Thickness t
Let plate separation be d, plate area be A, and original capacitance without dielectric be:
A dielectric slab of thickness t and dielectric constant K is inserted between the plates. The remaining air gap is:
The system behaves like two capacitors in series:
- Air region of thickness
d-t - Dielectric region of thickness
t
The effective separation becomes:
Therefore, capacitance is:
Using C0=ε0A/d, we can write:
So:
C' = ε0A/(d-t+t/K), or C' = C0d/(d-t+t/K).Special Case
If dielectric completely fills the space, then t=d:
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 p̄ |
-e |
Same mass, opposite charge and baryon number. |
Neutron n |
0 |
Antineutron n̄ |
0 |
Same mass but opposite baryon number and magnetic moment. |
Annihilation Example
Electron and positron can annihilate:
This process converts rest mass into electromagnetic radiation while conserving energy, momentum and charge.
Q8(b): Detection of Beta-Minus Radiation
Beta-minus radiation consists of high-speed electrons emitted from unstable nuclei. In beta-minus decay:
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
- Beta particle enters the GM tube through a thin window.
- It ionizes gas molecules.
- Electrons accelerate toward the anode wire.
- An avalanche of ionization occurs.
- 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. |
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:
Uncertainty principle states:
So:
Using ℏ = 1.055×10⁻³⁴ J s:
The corresponding relativistic energy would be approximately:
Convert to electron-volts:
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:
The electron is not a pre-existing nuclear constituent. It is produced at the moment of beta decay.
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. |
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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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