Use K.E. = (kq₁q₂)/r at closest approach

Convert 5MeV to joules and substitute charges

Calculate r ≈ 10⁻¹² cm

α particle changes mass by -4, atomic number by -2 β⁻ particle changes mass by 0, atomic number by +1

Track changes in mass number and atomic number

One α and two β gives stable isotope

Use conservation of momentum: m₁v₁ = m₂v₂ Mass ratio m₁:m₂ = r₁³:r₂³ = 1:8

v₁/v₂ = m₂/m₁ = 8/1

Therefore velocity ratio is 8:1

After first half-life: 50% remains Initial activity N₀/2

Final activity = 12.5% = N₀/8 Use N = N₀(1/2)^(t/T₁/₂)

Solve to get t = 40 minutes

For Lyman series, transitions end at n₁=1 Least energetic means starting from n₂=2

Use λ = 1/R(1/n₁² - 1/n₂²) = 1/0.0097(1/1² - 1/2²)

Calculate to get λ = 122 nm

Use ΔE = hν = (13.6 eV)(1/n₁² - 1/n₂²)

Substitute n₁=2, n₂=3

Simplify to get ν = 5RC/36

Convert KE to joules: 120×1.6×10⁻¹⁹ J

Calculate momentum: p = √(2mKE)

λ = h/p gives 112 pm

KE₁ = hν₁ - φ = 1 - 0.5 = 0.5 eV

KE₂ = hν₂ - φ = 2.5 - 0.5 = 2 eV

Since KE = ½mv², v₁:v₂ = √(0.5):√2 = 1:2

Apply conservation of momentum: pᵧ = P

Calculate energy using E = p²/2M

Convert units and solve for kinetic energy = 1 KeV

Use decay formula: N = N₀e^(-λt)

Calculate λ = ln2/T½ = ln2/10

Solve: N = 1000x×e^(-ln2×5/10) = 707x

Use quantization rule: mvr = nh/2π

Calculate difference: Δ(mvr) = (4-1)h/2π

Change in angular momentum = 3h/2π

Apply virial theorem: K = -E, U = 2E

Given E = -3.4 eV

Calculate: K = 3.4 eV, U = -6.8 eV

Use de Broglie formula: λ = h/p = h/√(2meV)

Substitute values in λ = 1.23/√V nm

Calculate: λ = 1.23/√400 = 0.06 nm

Convert mass defect to energy: E = mc² = 0.03 × 931 MeV

Total binding energy = 27.93 MeV

Per nucleon = 27.93/4 = 6.9825 MeV

Write nuclear equation

Check charge and mass conservation

Requires one alpha and two beta emissions

Consider alpha particle scattering experiment

Rutherford observed large angle scattering

This led to discovery of nuclear model

Use energy formula E = -13.6/n² eV

Calculate energies: E₂ = -13.6/4 eV, E₄ = -13.6/16 eV

Energy required = E₄ - E₂ = -13.6/16 - (-13.6/4) = +2.55 eV

Consider role of moderator

Moderator slows fast neutrons to thermal energies

Slower neutrons have higher fission cross-section

Use de Broglie relation: λ = h/mv

Calculate v using K.E = ½mv² = qV

Compare wavelengths considering mass and charge differences

Calculate number of half-lives: 50/12.5 = 4

Use formula: m = m₀(1/2)ⁿ where n=4

Calculate: 64 × (1/16) = 4 mg

Use λ = h/mv and v = √(2gH)

Substitute v in wavelength formula

Get λ ∝ 1/√H or H⁻¹/²

Convert 1g to kg: m = 10⁻³ kg

Use Einstein's equation: E = mc²

Calculate: E = 10⁻³ × (3×10⁸)² = 9×10¹³ J

Use T ∝ r³/² where r = n²a₀

For n=1 to n=2, ratio is (2²/1²)³/²

Calculate: T₂/T₁ = 8

Use Bohr's energy formula: E_n = -13.6/n² eV

For n=2, substitute in formula: E₂ = -13.6/2²

Calculate: E₂ = -13.6/4 = -3.4 eV

Use radiation pressure formula: P = F/A = n(h/λ)/t

Rearrange to solve for n: n = (Fλ/h)t

Calculate: n = (6.62×10⁻⁵ × 5×10⁻⁷)/(6.62×10⁻³⁴) = 5×10²²

Recall Einstein's photoelectric equation

Note that K.E_max = hf - φ

Observe dependence on frequency only

Nuclear force range ≈ 10⁻¹⁵ m

At 10nm = 10⁻⁸ m, nuclear force negligible

Only Coulomb force significant

First use frequency equation: f = qB/2πm to verify field strength (10⁶ Hz = [1.6×10⁻¹⁹ C × 0.66 T]/[2π × 1.67×10⁻²⁷ kg])

Use energy equation: E = q²B²R²/2m where R = 0.6m

Calculate final energy in MeV: E = [1.6×10⁻¹⁹)² × (0.66)² × (0.6)²]/[2 × 1.67×10⁻²⁷] = 7 MeV

Change in mass number = 232 - 208 = 24

Each α particle reduces mass by 4 units

Need 6α and 4β to match atomic number

Angular momentum in nth orbit = nh/2π

Change in L = Δ(nh/2π) = h/2π

Convert to Js: h/2π = 1.05x10⁻³⁴ Js

Use Bohr's model: v = √(ke²/mr)

r ∝ n², v ∝ 1/n

Frequency f = v/2πr ∝ 1/n³

λ = h/mv depends only on speed

Magnetic force is perpendicular to velocity

Speed remains constant, only direction changes

Impact parameter is perpendicular distance from initial path to target

For head-on collision, particle aims directly at nucleus

Therefore impact parameter = 0

Saturation current proportional to intensity

Frequency only affects electron energy

Double intensity means double current

In β⁻ decay, a neutron converts to a proton by weak interaction: n → p + e⁻ + ν̄ₑ

The electron is not present in nucleus before decay but is created during decay process along with antineutrino.

This explains charge and mass number conservation in β⁻ decay.

After one half-life (15 years), half of sample remains: N(15)/N₀ = 1/2. After two half-lives (30 years): N(30)/N₀ = (1/2)² = 1/4.

Fraction remaining = 1/4. Therefore, fraction decayed = 1 - 1/4 = 3/4 = 0.75.

This shows 75% of original sample has decayed in 30 years.

Nuclear force is strong but very short range (about 10⁻¹⁵ m). At 10 nm = 10⁻⁸ m, nuclear force is practically zero.

Coulomb force is long range and still significant at 10 nm. Calculate: Fₑ = kq²/r².

Therefore, Fₑ > Fₙ as electromagnetic force dominates at this large separation.

In Bohr model, angular momentum L = nh/2π where n is orbit number. Given L = 3h/2π, so n = 3.

For nth orbit, Total Energy = -13.6/n² eV and KE = +13.6/n² eV. For n = 3, calculate KE.

KE = 13.6/9 = 1.51 eV. Remember KE is positive and half the magnitude of total energy.

First, calculate photon energy at 600nm: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(600×10⁻⁹) = 2.07 eV

Compare this energy with band gaps: For detection, photon energy must exceed band gap. 2.07 eV is greater than D₂'s band gap (2eV) but less than D₁ (2.5eV) and D₃ (3eV).

Therefore, only D₂ can detect this light as it has sufficient energy to excite electrons across its band gap.

Use Bohr's model: r ∝ n², v ∝ 1/n. Period T = 2πr/v.

Substitute the relationships: T = 2π(kn²)/(k'/n) where k, k' are constants.

Simplify to get T ∝ n³. Period increases as cube of principal quantum number.

For ground state of hydrogen, E = -13.6 eV. Use de Broglie wavelength formula: λ = h/p = h/mv.

Find velocity from energy: ½mv² = 13.6 eV. Convert to appropriate units and solve for v.

Calculate wavelength using λ = h/mv. Result is 3.3 Å.

First understand that the electron starts with zero velocity and gains kinetic energy from the electric field. The work done by electric field (eV) equals the final kinetic energy (½mv²).

Carefully substitute values: Electric potential energy = eV = (1.6×10⁻¹⁹ C)(1200 V), mass of electron = 9.1×10⁻³¹ kg. Set this equal to ½mv².

Solve for velocity: v = √(2eV/m) = √(2×1.6×10⁻¹⁹×1200)/(9.1×10⁻³¹) = 2.1×10⁷ m/s

Let's understand the purpose of moderators first: they slow down fast neutrons to thermal energies for sustaining chain reaction.

In elastic collisions, maximum energy transfer occurs when masses are similar. For neutrons hitting heavy nuclei, very little energy is transferred per collision (like a ping pong ball hitting a bowling ball).

Light nuclei like hydrogen (in water) or carbon (in graphite) are much more effective moderators because their mass is closer to neutron mass, allowing more energy transfer per collision.

Use angular momentum quantization: mvr = 3h/2π. For Li²⁺, Z=3.

De Broglie wavelength λ = h/mv = h/(3h/2πr) = 2πr/3

Given r = a₀/Z where a₀ is Bohr radius. Substitute and solve to get P = 2.

For hydrogen-like atoms, En = -13.6(Z²/n²) eV. For hydrogen in n=2, E₂ = -13.6/4 eV.

For He⁺, Z=2 and n=3. E = -13.6(4/9) eV

Taking ratio of He⁺(n=3) to H(n=2): (-13.6×4/9)/(-13.6/4) = 16/9

First, calculate photon energy: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(3000×10⁻¹⁰) = 4.14 eV

Using Einstein's photoelectric equation: KE = hν - φ = 4.14 - 1 = 3.14 eV. Convert to Joules: 3.14×1.6×10⁻¹⁹ J

For velocity: ½mv² = KE. Solving for v = √(2KE/m) ≈ 10⁶ m/s

Consider particle nature and charge of each: γ-rays (EM waves), α-particles (He²⁺), X-rays (EM waves), neutrons (neutral)

Only charged particles are deflected by electric field

α-particles are positively charged, so they're deflected! Others aren't affected

Use E = mc². Power is energy per time, so P = (mc²)/t

Given P = 9×10⁹W = 9×10⁹J/s, t = 3600s (1 hour)

Solve for m: m = (Pt)/(c²) = 0.04g. Small mass produces huge energy!

Use Einstein's photoelectric equation: KE = hc(1/λ₂ - 1/λ₁)

Given ΔKE = 0.52eV = hc(1/λ₂ - 1/500nm)

Solve for λ₂, getting 400nm. Shorter wavelength means higher energy!

In Bohr model, radius of nth orbit is r_n = n²r₁, where r₁ is ground state radius

For n=3: r₃ = (3²)(0.529Å) = (9)(0.529Å)

Calculate: r₃ = 4.761Å. Keep units consistent!

Saturation current depends on number of electrons emitted per second

Intensity determines number of photoelectrons (doubled), frequency affects KE

When both double, only intensity affects current, so current doubles!

Binding energy = [Zm_p + (A-Z)m_n - M]c² where M is actual mass

For ¹⁴N: Z=7, A=14. Calculate mass defect using given mass

Convert to energy using E=mc². BE = 104.7 MeV. Check units!

de Broglie wavelength λ = h/p = h/√(2mK). Let's write this for both energies

For K/4: λ₂ = h/√(2m(K/4)) = h/√(mK/2)

Compare with original λ₁ = h/√(2mK): λ₂/λ₁ = √2 = 2. So λ₂ = 2λ₁!

In Bohr model, radius ∝ n². Let's set up ratio: r₂/r₁ = (n₂/n₁)²

Given r₁ = 0.53Å (n₁=1), r₂ = 2.12Å. Substitute: 2.12/0.53 = n₂²

Solve: 4 = n₂², therefore n₂ = 2. The electron jumped to second orbit!

Use N=N₀e⁻λt

For 1/5th: 1/5=e⁻λt

t=(ln5)/λ≈7 years

Convert energy to joules: E=2.4×1.6×10⁻¹⁹J

Use p=E/c for photon momentum

p=1.28×10⁻²⁷ kg.m.s⁻¹

Use momentum conservation

KE=(A-4/A)Q

KE=(216/220)×5.5=5.4MeV

Write nuclear equation

Calculate neutrons using mass number

n=A-Z=210-84=126

Calculate energy per photon: E=hc/λ

Find number using P=nE/t

n=Pt/E=2×10²⁰ photons/s

Use Einstein's equation: hc/λ=φ

Calculate threshold wavelength: λ₀=12400/2.4=5166Å

700nm>516.6nm, no emission

Large impact parameter means particle passes farther

Force decreases with distance

Scattering angle decreases

Apply photoelectric equation: E=φ+KEmax

Substitute φ=E/3 to get KEmax=2E/3

KE range is 0 to 2E/3

Calculate total binding energy difference: (1980-1800)MeV

Energy per neutron=(180)/3 MeV

Each neutron carries 60 MeV

Use Bohr's theory: radius of orbit r∝n² where n is principal quantum number

Area of orbit ∝r² ∝n⁴

Compare n=2 (first excited) to n=1 (ground): (2⁴:1⁴)=16:1

Use R=R₀A¹/³ for nuclear radius

Calculate ratio: V/A=(4πR³/3)/(4πR²)=R/3

Substitute R=R₀(27)¹/³=1.2×10⁻¹⁵m

Use relationship d∝1/v² from Rutherford formula

When v→2v, d→d/4

New distance is d/4