#physics formula

Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.25 \times 10^8 \, \text{m/s} \) and in vacu

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.25 × 10⁸ m/s . n = (3.0 × 10⁸/2.25 × 10⁸) = 1.33 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

Ref: NCERT > Physics Book > Wave Optics > Refraction, Refractive Index and Critical Angle

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Radioactive Decay, Nuclear Forces and Stability

A nucleus with mass number 216 has a radius of \( 7.2 \times 10^{-15} \, \text{m} \). What is the value of \( R_0 \)?

**Radioactive decay** occurs when nucleus unstable, alpha decay emits He-4, beta decay neutron→proton+electron+antineutrino, gamma decay photon emission, decay law N=N₀ e^{-λt}, half-life T½=ln2/λ, nuclear density ~10¹⁷ kg/m³, nuclear force saturated means BE/A constant for A>20. R = R₀ A¹/³ . A = 216 , A¹/³ = 6 . R₀ = (R/A¹/³) = (7.2 × 10⁻¹⁵/6) = 1.2 × 10⁻¹⁵ m . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 1.2 × 10⁻¹⁵ m, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Mass Defect, Binding Energy and Binding Energy per Nucleon

InEinstein's mass-energy equivalence, what does the term \( c^2 \) represent?

**Binding energy calculation** from mass defect, for nucleus mass number 28 BE 224 MeV BE/A=8 MeV, A=18 BE 144 MeV BE/A=8 MeV, BE/A indicates stability, fusion of light nuclei and fission of heavy release energy because product has higher BE/A, difference released. In E = m c² , c² is the square of the speed of light in a vacuum, acting as the conversion factor between mass and energy, indicating the immense energy equivalent of a small mass. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV,

Ref: NCERT > Physics Book > Atoms and Nuclei > Mass Defect, Binding Energy and Binding Energy per Nucleon

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Ultraviolet, X-rays, Gamma Rays and Their Properties

An electromagnetic wave has a wave number \( k = 4 \, \text{rad/m} \). What is its wavelength in vacuum?

**Gamma rays** λ10¹⁹ Hz, produced by nuclear transitions and radioactive decay, associated with nuclear processes because nuclear energy levels MeV vs atomic eV, highest photon energy, used in cancer treatment due to high penetration and cell damage, also sterilization, astronomy. The wave number k = (2 π/λ) . Given k = 4 rad/m , we have λ = (2 π/k) = (2 π/4) ≈ 1.57 m . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.57 m, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Ultraviolet, X-rays, Gamma Rays and Their Properties

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material with susceptibility \( \chi = 2 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

**Relation between B, H, M** is B = μ₀(H+M) = μ₀(1+χ)H. Susceptibility χ = μ_r -1 quantifies material response. Given B, μ_r, n, current I = B/(μ₀ μ_r n), with μ₀ = 4π×10⁻⁷ T·m/A, enabling current calculation for desired B with magnetic core. μ_r = 1 + chi . Given: chi = 2 × 10⁻³ . Substitute: μ_r = 1 + 2 × 10⁻³ = 1.002 . Substituting values gives 1.002, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Magnetization, Magnetic Intensity, Susceptibility and Permeability

A material has a magnetic susceptibility \( \chi = 0.001 \). What is its relative magnetic permeability \( \mu_r \)?

**Magnetic properties** μ_r = 400 indicates 400 times vacuum permeability, so B enhanced 400 times for same nI. H = nI (A/m) for solenoid, M = χ H, B = μ₀(H+M) links microscopic magnetization to macroscopic field. Relative magnetic permeability μ_r = 1 + chi . Given: chi = 0.001 . Substitute: μ_r = 1 + 0.001 = 1.001 . Substituting values gives 1.001, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetization, Magnetic Intensity, Susceptibility and Permeability

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Escape Velocity and Bound Systems

What is the significance of the negative sign in the gravitational potential energy formula V\=−GMmr?

The negative sign indicates that gravitational potential energy is zero at infinite separation (r→∞) and decreases (becomes more negative) as objects approach each other. It reflects the attractive nature of gravity, where work must be done to separate masses.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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