Alpha particle charge = +2e. Kinetic energy after acceleration = 2eV. At closest approach KE = Coulomb PE = \dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(Ze)}{r}. Solving for r gives r = \dfrac{Ze^2}{4\pi\epsilon_0 eV}. Using e^2/(4\pi\epsilon_0)=14.4 eV·Å, r = 14.4\,\dfrac{Z}{V}\,Å.
Bohr quantization: $L = n\dfrac{h}{2\pi}=n\hbar$. For n=4, $L=4\dfrac{h}{2\pi}=\dfrac{4h}{2\pi}=\dfrac{2h}{\pi}$.
Ionization potential for hydrogenic (n=1) is $\mathrm{IP}=13.6\,Z^2\,\text{eV}$. So $Z^2=122.4/13.6=9\Rightarrow Z=3$.
Bohr radius $r_n\propto n^2$. Thus $r_1:r_2:r_3=1^2:2^2:3^2=1:4:9$.
Cathode rays are streams of electrons which carry negative charge.
For undeflected motion $E=vB\Rightarrow v=E/B=3.0\times10^4/(1.0\times10^{-3})=3.0\times10^7\,$m/s. From acceleration $\tfrac12 mv^2=eV\Rightarrow e/m=v^2/(2V)$. So $e/m=(3.0\times10^7)^2/(2\times2600)\approx1.73\times10^{11}\,$Ckg^{-1}.
For hydrogen-like ions, wavelength $\lambda\propto1/Z^2$. For Li^{2+}(Z=3), He^{+}(Z=2), H(Z=1): ratios $\propto1/9:1/4:1=4:9:36$ after clearing denominators.
Force from potential $F=-e\,dV/dr\propto -1/r$. Thus central force ~1/r. For circular motion $m v^2/r\propto1/r\Rightarrow m v^2=$ constant so $v$ independent of r. Bohr quantization $mvr=n\hbar$ gives $r\propto n$.
Focal length of a thin plano-convex lens: $\dfrac{1}{f}= (n-1)(1/R)$. With R=10 cm, n=1.5 gives $f_{lens}=20\,$cm. For silvered plano-convex (plane face silvered), effective focal length becomes $f=\dfrac{R}{2(n-1)}=\dfrac{10}{2\times0.5}=10\,$cm.
Let real depth from one face be d. Apparent depths: d/\mu=5 and (t-d)/\mu=3. Adding gives t/\mu=8 so $t=8\mu=8\times1.5=12\,$cm.
Answer (derivation):
From kinetic theory for an ideal gas: p = (1/3)(N/V) m <v^2> , so
pV = (1/3) N m <v^2> .
For a gas at a given temperature, the quantity m <v^2> (and hence the factor relating pressure, number and volume) is fixed. Therefore, at fixed pressure p and temperature T, V ∝ N. Thus equal volumes of gases at the same temperature and pressure contain equal numbers of molecules — this is Avogadro's law.
Each α reduces A by 4 and Z by 2. Two α: A' = A-8, Z' = Z-4. Two positron emissions reduce Z' by additional 2 (each p→n), so final Z_f=Z-6 and final N_f = A_f - Z_f = (A-8)-(Z-6)=A-Z-2. So neutron/proton = (A-Z-2)/(Z-6).
Half-life T_{1/2}=\ln2/\lambda and mean life \tau=1/\lambda. Given T_{1/2}(A)=\tau(B) so \lambda_A=\ln2/\tau_B and \lambda_B=1/\tau_B. Thus \lambda_B/\lambda_A=1/(\ln2)\approx1.44>1 so B has larger decay constant and decays faster initially.
After time t, $N=N_0(1/2)^{t/T_{1/2}}$. For $t=\tfrac{1}{2}T_{1/2}$, $N=N_0(1/2)^{1/2}=N_0/\sqrt{2}\approx0.707N_0$. (None of the given options equals this value.)
Observed in vacuum tubes, cathode rays are negatively charged particles (electrons) produced at the cathode and accelerated toward the anode.
Cathode rays are streams of electrons emitted from the cathode in a discharge tube when high voltage is applied; they travel in straight lines and cause fluorescence.
Experimental observations (deflection by fields, shadow casting, fluorescence) establish these properties.
Properties: (i) Travel in straight lines, (ii) made of negatively charged particles (deflected by E and B fields toward positive), (iii) produce fluorescence on striking glass, (iv) cause heating and phosphorescence, (v) have particle nature with charge-to-mass ratio e/m.
Observed angular distribution led to nuclear model: tiny, dense, positively charged nucleus with electrons outside.
Results: (i) Most α-particles pass through with little deflection → atom mostly empty space; (ii) a few are deflected by large angles → concentrated positive charge in a small nucleus; (iii) some are back-scattered indicating nucleus is heavy and compact.
These give discrete energy levels $E_n=-13.6\,Z^2/n^2\,$eV for hydrogen-like atoms.
Main postulates: (i) Electrons move in circular orbits under Coulomb force without radiating energy (stationary states); (ii) Angular momentum quantization: $mvr=n\hbar$, $n=1,2,\dots$; (iii) Radiation emitted/absorbed when electron jumps between orbits with $\Delta E= h\nu$ equal to energy difference.
For hydrogenic atom, excitation energy from ground to level n is $E_n-E_1=13.6Z^2(1-1/n^2)\,$eV.
Excitation energy is the energy required to raise an electron from a lower energy level to a higher (excited) level in an atom.
For hydrogen ground state, ionization energy =13.6 eV and ionization potential =13.6 V.
Ionization energy: energy required to remove an electron from the atom (in eV or J). Ionization potential: the corresponding potential in volts needed to remove electron (1 eV corresponds to 1 V for single electron).
Bohr's model was superseded by quantum mechanics which handles these phenomena.
Drawbacks: (i) Applies only to hydrogen-like atoms (single electron); (ii) Cannot explain fine structure, Zeeman splitting, or spectral intensities; (iii) Contradicts Heisenberg uncertainty principle; (iv) No prediction of multi-electron interactions.
Computed by equating initial KE to electrostatic potential energy at that distance.
Distance of closest approach is the minimum separation between an incident charged particle (e.g., α-particle) and the nucleus when the particle momentarily comes to rest (all KE converted to Coulomb PE).
Used in scattering theory to relate scattering angle to closest approach.
Impact parameter is the perpendicular distance between the initial velocity vector of an incoming particle (if undeflected) and the centre of the target nucleus.
Standard nuclear notation indicates composition and identity of nucleus.
Notation: ^A_ZX where A is mass number (protons+neutrons), Z is atomic number (number of protons), and X is chemical symbol. Number of neutrons = A-Z.
They have identical chemical properties but different masses and nuclear properties.
Isotopes are nuclei of the same element (same Z) with different mass numbers A (different neutrons). Example: ^12C and ^14C are isotopes of carbon.
Isotones share neutron count but differ chemically.
Isotones are nuclei with the same neutron number N but different proton numbers Z. Example: ^14C (Z=6,N=8) and ^15N (Z=7,N=8).
Isobars have same total nucleon number but different compositions of protons and neutrons.
Isobars are nuclei with the same mass number A but different Z. Example: ^14C (Z=6,A=14) and ^14N (Z=7,A=14).
Common mass unit used in nuclear and atomic physics; energy equivalent $1\,u\approx931.494\,$MeV.
Atomic mass unit (u) is defined as one twelfth of the mass of a neutral ^12C atom: $1\,u=1/12\,m(^{12}\mathrm{C})\approx1.66054\times10^{-27}\,$kg.
Mass~Am_u and volume~(4/3)\pi R_0^3 A so cancels A leading to nearly constant density independent of A.
Using nuclear radius $R=R_0A^{1/3}$ with $R_0\approx1.2\,$fm, nuclear volume $\propto A$, so density $\rho=\dfrac{mass}{volume}\propto\dfrac{A}{A}=\text{constant}$. Numerically $\rho\approx2.3\times10^{17}\,\mathrm{kg\,m^{-3}}$.
Mass defect times $c^2$ equals the binding energy of the nucleus.
Mass defect is the difference between the sum of masses of separate nucleons and the actual mass of the nucleus: $\Delta m=Zm_p+Nm_n - m_{nucleus}$.
It is a measure of nuclear stability; binding energy per nucleon is B/A.
Binding energy B is the energy required to disassemble a nucleus into separate protons and neutrons. $B=\Delta m\,c^2=(Zm_p+Nm_n - m_{nucleus})c^2$.
Using $1\,u=1.66054\times10^{-27}\,$kg and $E=mc^2$ gives $E\approx1.66054\times10^{-27}\times(3.0\times10^8)^2\,$J then convert to MeV.
Energy equivalent: $1\,u\,c^2\approx931.494\,$MeV $\approx1.492\times10^{-10}\,$J.
Plot of $B/A$ vs A shows maximum around iron (A~56) indicating most stable nuclei.
Binding energy per nucleon $B/A$ measures average energy that would be required to remove one nucleon from the nucleus. It indicates nuclear stability; higher $B/A$ means more stable nucleus.
It follows exponential decay laws characterized by decay constant, half-life and mean life.
Radioactivity is the spontaneous transformation of an unstable nucleus into a more stable one accompanied by emission of particles (α, β) and/or γ radiation.
Symbols show conservation of nucleon number and appropriate changes in Z.
Alpha: ^A_ZX → ^{A-4}_{Z-2}Y + ^4_2He. Beta^-: ^A_ZX → ^A_{Z+1}Y + e^- + \bar\nu_e. Beta^+: ^A_ZX → ^A_{Z-1}Y + e^+ + \nu_e. Gamma: ^A_ZX^* → ^A_ZX + γ.
Quantum-mechanical formation probability and lower Coulomb barrier for clustered emission explain α-particle emission.
Emission of a pre-formed tightly bound α-particle (with high binding energy) is energetically favourable and requires lower barrier penetration than emitting four separate nucleons. The α cluster is compact and more likely to tunnel out collectively.
Relation to half-life: $T_{1/2}=\tau\ln2$.
Mean life $\tau$ is average lifetime of a nucleus before decay: $\tau=1/\lambda$ where $\lambda$ is decay constant.
From $N=N_0e^{-\lambda t}$ set $N=N_0/2$ to obtain the expression.
Half-life $T_{1/2}$ is the time required for half the nuclei to decay. $T_{1/2}=\dfrac{\ln2}{\lambda}$.
Activity decreases with time as $A(t)=A_0e^{-\lambda t}$.
Activity A is the number of decays per unit time: $A=\lambda N$. SI unit is becquerel (Bq) = s^{-1}; older unit curie (Ci) where 1 Ci = 3.7×10^{10} s^{-1}.
Approximately the activity of 1 gram of radium-226.
Curie (Ci) is an older unit of radioactivity defined as 3.7×10^{10} decays per second.
Quark model explains charges and other properties: u-charge +2/3, d-charge -1/3 giving proton charge +1 and neutron 0.
Proton and neutron are each made of three quarks: proton = (uud), neutron = (udd), bound by gluons in quantum chromodynamics.
Concise derivation: (i) Balance E and B: eE = evB ⇒ v = E/B. (ii) Magnetic deflection: evB' = mv^2/R ⇒ e/m = v/(B'R). Using measured R and B' yields e/m. Alternatively for acceleration through V: (1/2)mv^2=eV ⇒ e/m=v^2/(2V).
J.J. Thomson measured e/m by deflecting cathode rays with crossed electric and magnetic fields. By balancing fields (no deflection) he obtained velocity v=E/B. Then by removing E and measuring magnetic deflection radius R, centripetal force mv^2/R=e v B' gave e/m=v/(B'R). Combining gives e/m=v^2/(2V) if accelerated through known potential V. This yields the specific charge e/m.
Key relations: with no field, viscous drag equals weight minus buoyancy to find radius via Stokes' law. With field, qE balances gravity corrected by buoyancy and drag. Solve for q, compile q values to deduce smallest common divisor e≈1.602×10^{-19} C.
Millikan balanced gravitational and electric forces on charged oil drops between capacitor plates. Measuring terminal velocities with and without electric field gave drop radius and mass, and from equilibrium with field he obtained the drop charge. Repeating for many drops revealed charges always integer multiples of a fundamental charge e, yielding value of electron charge.
Show stepwise algebra eliminating v and r using quantization to arrive at $E_n=-13.6\,Z^2/n^2$ eV for hydrogen-like ions.
Using Coulomb force and quantization: $\dfrac{ke^2}{r^2}=m v^2/r$ and $mvr=n\hbar$. Solve for r_n and v_n to get $r_n=\dfrac{4\pi\epsilon_0\hbar^2}{m e^2}n^2=a_0 n^2$. Total energy $E_n=K+U=\tfrac12 mv^2 - \dfrac{ke^2}{r}=-\dfrac{ke^2}{2r_n}=-\dfrac{m e^4}{8\epsilon_0^2 h^2}\dfrac{1}{n^2}=-\dfrac{13.6\,\mathrm{eV}}{n^2}$ for hydrogen (Z=1).
Describe energy level diagram and examples of prominent lines (e.g., Balmer H_α: n=3→2 at 656 nm).
Spectral series correspond to transitions terminating at a particular principal quantum number n_f: Lyman (n_f=1, UV), Balmer (n_f=2, visible), Paschen (n_f=3, IR), Brackett (n_f=4), Pfund (n_f=5). Wavelengths given by Rydberg formula $1/\lambda=R_H(1/n_f^2-1/n_i^2)$ with $n_i>n_f$.
Discuss contributions: volume term, surface, Coulomb repulsion, asymmetry, pairing (semi-empirical mass formula) explain curve shape and maximum.
Plot of binding energy per nucleon B/A vs A: rises steeply for small A, reaches a maximum around A~56 (Fe), then slowly decreases for heavy nuclei. Features: (i) Light nuclei gain energy via fusion (move to higher B/A); (ii) Heavy nuclei can release energy by fission (splitting increases B/A); (iii) local irregularities due to shell effects; (iv) explains source of nuclear energy in stars and reactors.
Mention Yukawa potential $V(r)\sim -g^2\dfrac{e^{-\mu r}}{r}$ and experimental evidence: short range, binding energies, existence of deuteron and absence of nn bound state, scattering experiments.
Nuclear (strong) force: short-range (effective up to ~2–3 fm), attractive at intermediate distances binding nucleons, strongly repulsive at very short ranges (<~0.5 fm) to prevent collapse, charge-independent (similar for pp, nn, np), has spin and isospin dependence, and saturates (each nucleon interacts only with nearest neighbors). It is mediated by mesons (Yukawa theory) at low energies and by gluons between quarks in QCD.
Describe Gamow theory: decay constant λ ~ frequency × tunnelling probability (exponential dependence on Q and Z), explains range of half-lives across isotopes.
Alpha decay: nucleus emits an α-particle (^4_2He), reducing mass and charge. Example: ^{238}_{92}U → ^{234}_{90}Th + ^4_2He. Decay explained by quantum tunnelling through Coulomb barrier; decay rate determined by barrier height/width and pre-formation probability. Energy release (Q-value) equals mass difference times c^2 and is shared as kinetic energy (mostly to daughter nucleus recoil).
Explain conservation laws (energy, momentum, charge, lepton number), role of neutrino (Pauli proposed neutrino to account for continuous electron spectrum), Fermi theory of beta decay.
Beta decay involves transformation of a neutron to proton (β^- emission) or proton to neutron (β^+ emission or electron capture). Examples: β^-: ^{14}_6C → ^{14}_7N + e^- + \bar\nu_e. β^+: ^{11}_6C → ^{11}_5B + e^+ + \nu_e. Energy shared continuously among emitted electron/positron and (anti)neutrino. Weak interaction governs beta decay.
Discuss selection rules, internal conversion (competing process), and detection methods (Geiger–Müller, scintillators, HPGe detectors).
Gamma emission: de-excitation of nucleus from excited state to lower state emitting a photon (γ) with no change in A or Z. Example: ^{60}_{27}Co* → ^{60}_{27}Co + γ. Gammas often follow α or β decays when daughter nucleus is left in excited state. Photons have high energy (keV–MeV).
Derivation from first-order differential equation and definitions follows standard calculus.
Radioactive decay law: the rate of decay is proportional to number of undecayed nuclei: $\dfrac{dN}{dt}=-\lambda N$. Integrating: $N(t)=N_0 e^{-\lambda t}$. Activity $A(t)=\lambda N(t)=A_0 e^{-\lambda t}$. Half-life $T_{1/2}=\ln2/\lambda$ and mean life $\tau=1/\lambda$.
Pauli postulated neutrino; Fermi formalism incorporated it. Detection requires large detectors because of tiny cross-section.
Neutrino: neutral, very small (possibly zero) mass, weakly interacting, comes in three flavours (ν_e, ν_μ, ν_τ), spin-1/2 fermion. In β^- decay a (anti)neutrino carries away missing energy and momentum, conserving energy and angular momentum and resolving continuous electron spectrum puzzle.
Assume initial ratio equals modern atmospheric ratio; correction for calibration and reservoir effects may be needed for accuracy.
Carbon dating uses decay of ^14C (half-life ~5730 yr). Living organisms have constant ^14C/^12C ratio from atmosphere. After death, ^14C decays. Measuring remaining ^14C activity gives age via $t=\dfrac{\ln(N_0/N)}{\lambda}$ where λ=\ln2/T_{1/2}.
Discuss criticality (subcritical, critical, supercritical), role of moderators, control rods, and issues of radioactivity and waste.
Nuclear fission: heavy nucleus splits into two lighter nuclei plus neutrons, releasing energy (from increased binding energy per nucleon). Triggered by neutron absorption (e.g., ^{235}U + n → fission fragments + 2–3 n + ~200 MeV). Properties: chain reaction possible, moderated by neutron economy, produces fission products and neutrons, large energy per event, used in reactors and bombs.
Describe proton–proton chain steps, energy outputs, and significance for stellar life cycles.
Nuclear fusion: light nuclei combine to form heavier nuclei, releasing energy when products have higher binding energy per nucleon (e.g., in stars H→He via proton–proton chain or CNO cycle). High temperature and pressure overcome Coulomb barrier, allowing quantum tunnelling. In stars, fusion releases enormous energy maintaining hydrostatic equilibrium and producing stellar radiation.
Explain roles of each component and safety systems (containment, control rods, emergency shutdown), plus moderator examples (graphite, heavy water).
Reactor components: fuel (e.g., ^{235}U), moderator (slow neutrons), control rods (absorb neutrons to control rate), coolant (remove heat), pressure vessel, steam generator and turbine for electricity. Chain reaction sustained at criticality produces heat used to produce steam driving turbines.
Mention relative strengths, ranges and role in particle interactions and unification attempts (electroweak, grand unification).
Four forces: (i) Gravitational — universal attraction, long-range, weakest, responsible for large-scale structure; (ii) Electromagnetic — between charges, long-range, mediates chemistry and light; (iii) Weak nuclear — short-range, mediates β-decay and flavour change, responsible for radioactive decay; (iv) Strong nuclear — strongest at short range, binds quarks into nucleons and nucleons into nuclei. Each has characteristic carriers: graviton (hypothetical), photon, W±/Z^0, gluons.
Note conservation laws (charge, baryon number, lepton number) and interactions via gauge symmetries SU(3)×SU(2)×U(1).
Standard Model particles: Fermions — quarks (u,d,c,s,t,b) and leptons (e,μ,τ and corresponding neutrinos) in three generations; Bosons — gauge bosons (photon, W±, Z^0, gluons) mediating forces; scalar Higgs boson gives mass via Higgs mechanism. Quarks combine to form hadrons (baryons, mesons).
Minimum energy needed to excite one atom to first excited state (energy difference from ground to n=2 is 10.2 eV for hydrogen). In CM frame for perfectly inelastic head-on collision where two identical masses stick together, kinetic energy of moving atom must supply both excitation (10.2 eV) and provide recoil kinetic energy of combined mass. For minimum KE, all excess goes into excitation and center-of-mass motion minimized. Using energy and momentum conservation for equal masses colliding and sticking, fraction of initial KE available for internal excitation is 1/2. Therefore required initial KE = 2×10.2 eV = 20.4 eV.
20.4 eV
Compute: ν_{3→2}=R c (1/2^2 - 1/3^2)=Rc(1/4-1/9), ν_{2→1}=Rc(1-1/4), sum = Rc(1-1/9)=Rc(1-1/9)=ν_{3→1}. Thus Ritz combination principle holds.
ν_{3→2} + ν_{2→1} = ν_{3→1}
(a) Photon energy E=hc/λ = (1240 eV·nm)/97.5 nm ≈12.72 eV. Energy difference from ground to level n is 13.6(1-1/n^2). Solve 13.6(1-1/n^2)=12.72 ⇒1/n^2=1-12.72/13.6≈1-0.9353=0.0647 ⇒n^2≈15.45 ⇒n≈3.93 ≈4. (b) Number of possible emission lines from level n to lower levels = number of distinct pairs i<j with i,j ≤ n which equals n(n-1)/2. For n=4 gives 4×3/2=6 lines.
(a) n=4; (b) 6 lines
Take nuclear density ρ_n ≈ 2.3×10^{17} kg m^{-3}. Volume V = M/ρ_n = 5.97×10^{24}/2.3×10^{17}=2.595×10^{7} m^3. Radius r=(3V/4π)^{1/3}≈(3×2.595×10^7/4π)^{1/3}≈(6.21×10^6)^{1/3}≈181 m (≈180 m).
≈180 m
Compute: A=108, Z=47 so N=61. Sum masses of nucleons = 47 m_p + 61 m_n ≈ 47(1.00727647 u)+61(1.0086649 u)=47.341+61.528=108.869 u (approx). Nuclear mass given =107.905949 u. Δm ≈108.869-107.905949≈0.963051 u (numeric discrepancy due to used precise m_p,m_n). Using 1u=931.494 MeV gives binding energy B=Δm c^2 ≈0.963×931.494≈897.3 MeV. Binding energy per nucleon ≈897.3/108≈8.31 MeV. (Book quoted Δm≈0.99039185 u and correspondingly BE per nucleon ≈8.55 MeV depending on constants chosen).
Mass defect Δm ≈ 0.99039185 u; binding energy per nucleon ≈ (value given in book) MeV per nucleon
Initial N_0 for both. Remaining after t: N_A=N_0(1/2)^{80/20}=N_0(1/2)^4=N_0/16 so decayed ΔN_A=N_0-N_A=15N_0/16. For B: N_B=N_0(1/2)^{80/40}=N_0(1/2)^2=N_0/4 so ΔN_B=3N_0/4=12N_0/16. Ratio ΔN_A:ΔN_B=15:12=5:4.
5 : 4
(a) Time interval ≈1 year≈365 days. Activity decays: A=A_0 e^{-\lambda t} with λ=\ln2/T_{1/2}=\ln2/(5.01 d)=0.1384 d^{-1}=1.602×10^{-6} s^{-1}. Over 365 d, A/A_0= e^{-0.1384×365}=e^{-50.5}≈1.2×10^{-22}. So A≈1.2×10^{-22} μCi (≈10^{-22} μCi). (b) λ=\ln2/(5.01×86400)≈1.60×10^{-6} s^{-1}. (c) Mean life τ=1/λ≈6.24×10^5 s ≈7.23 days. (d) Initial activity 1 μCi =1×10^{-6} Ci = (1×10^{-6})(3.7×10^{10} s^{-1})=3.7×10^{4} s^{-1}. Number of atoms N_0=A_0/λ=3.7×10^4 /1.60×10^{-6} ≈2.31×10^{10}.
See solution
If 60% decayed, remaining fraction = 0.40 = e^{-\lambda t} where λ=\ln2/T_{1/2}=\ln2/3.8 d^{-1}=0.18245 d^{-1}. So t= -\ln(0.40)/λ = 0.91629/0.18245 ≈5.02 days.
t ≈ 5.02 days
1 watt =1 J/s. Energy per fission =200 MeV =200×10^6×1.602×10^{-19} J =3.204×10^{-11} J. Number per second =1 / (3.204×10^{-11}) ≈3.12×10^{10} fissions/s.
≈3.125×10^{10} fissions/s
1 Ci =3.7×10^{10} decays/s = A. Decay constant λ=\ln2/T_{1/2}=0.693/(1600×365×86400)≈1.37×10^{-11} s^{-1}. Number of atoms N=A/λ≈3.7×10^{10}/1.37×10^{-11}≈2.7×10^{21} atoms. Moles = N/ N_A ≈2.7×10^{21}/6.02×10^{23}≈4.5×10^{-3} mol. Mass = moles × atomic mass (226 g/mol) ≈4.5×10^{-3}×226≈1.02 g.
≈1 g
Remaining fraction =0.175 = e^{-\lambda t} with λ=\ln2/T_{1/2} and T_{1/2}(^{14}C)=5730 yr. So t= -\ln(0.175)/λ = -\ln(0.175)×T_{1/2}/\ln2 ≈1.742/0.693×5730 ≈2.513×5730 ≈14400 yr ≈1.44×10^4 yr.
≈1.44×10^4 years
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