Part of PH-02 — Atoms & Nuclei

Key Points (~300 words): Formulas with LaTeX

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Key Points: All Formulas with Dimensional Analysis

Rutherford: d=2kZe2KEα[d]=[L]=md = \frac{2kZe^2}{KE_\alpha} \quad [d] = [L] = \text{m}

Bohr Model: rn=0.529n2Z A˚[r]=[L]=mr_n = \frac{0.529\, n^2}{Z}\ \text{Å} \quad [r] = [L] = \text{m}

vn=2.18×106Zn m/s[v]=[LT1]v_n = \frac{2.18 \times 10^6 \cdot Z}{n}\ \text{m/s} \quad [v] = [LT^{-1}]

En=13.6Z2n2 eV[E]=[ML2T2]E_n = -\frac{13.6 Z^2}{n^2}\ \text{eV} \quad [E] = [ML^2T^{-2}]

KEn=En(positive)PEn=2En(negative)KE_n = -E_n \quad (\text{positive}) \qquad PE_n = 2E_n \quad (\text{negative})

Ln=n=nh2π[L]=[ML2T1]=J⋅sL_n = n\hbar = \frac{nh}{2\pi} \quad [L] = [ML^2T^{-1}] = \text{J·s}

Spectral Series: 1λ=RZ2(1n121n22),R=1.097×107 m1 [L1]\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad R = 1.097 \times 10^7\ \text{m}^{-1}\ [L^{-1}]

Nlines=n(n1)2N_{\text{lines}} = \frac{n(n-1)}{2}

Nuclear Physics: R=R0A1/3, R0=1.2 fm[R]=[L]R = R_0 A^{1/3},\ R_0 = 1.2\ \text{fm} \quad [R] = [L]

ρ=Amu(4/3)πR32.3×1017 kg/m3[ML3]=constant\rho = \frac{Am_u}{(4/3)\pi R^3} \approx 2.3 \times 10^{17}\ \text{kg/m}^3 \quad [ML^{-3}] = \text{constant}

Δm=[Zmp+(AZ)mn]M[Δm]=[M]=u or kg\Delta m = [Zm_p + (A-Z)m_n] - M \quad [\Delta m] = [M] = \text{u or kg}

BE=Δm×931.5 MeV[ML2T2](1 u=931.5 MeV/c2)BE = \Delta m \times 931.5\ \text{MeV} \quad [ML^2T^{-2}] \quad (1\text{ u} = 931.5\ \text{MeV}/c^2)

Radioactive Decay: N(t)=N0eλt=N02t/t1/2[λ]=[T1]=s1N(t) = N_0 e^{-\lambda t} = \frac{N_0}{2^{t/t_{1/2}}} \quad [\lambda] = [T^{-1}] = \text{s}^{-1}

t1/2=ln2λ=0.693λ[T]=st_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \quad [T] = \text{s}

τ=1λ=t1/20.693=1.443t1/2[T]=s\tau = \frac{1}{\lambda} = \frac{t_{1/2}}{0.693} = 1.443\, t_{1/2} \quad [T] = \text{s}

A(t)=λN(t)=A0eλt[T1]=BqA(t) = \lambda N(t) = A_0 e^{-\lambda t} \quad [T^{-1}] = \text{Bq}

Distance of Closest Approach: d=2kZe2KEα, k=9×109 Nm2C2, e=1.6×1019 Cd = \frac{2kZe^2}{KE_\alpha},\ k = 9 \times 10^9\ \text{Nm}^2\text{C}^{-2},\ e = 1.6 \times 10^{-19}\ \text{C}

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