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CalibratePhysics

Chapter 15

Structure of Atoms and Nuclei

Maharashtra HSC · 11 subtopics · textbook pages 324–343

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants and standard values to quote

  • Planck's constant h=6.626×10−34 J sh = 6.626\times10^{-34}\ \text{J s} (board papers often print 6.63×10−34 J s6.63\times10^{-34}\ \text{J s}).
  • Electronic charge e=1.602×10−19 Ce = 1.602\times10^{-19}\ \text{C}, and 1 eV=1.6×10−19 J1\ \text{eV} = 1.6\times10^{-19}\ \text{J}.
  • Permittivity of free space ε0=8.854×10−12 C2 N−1m−2\varepsilon_0 = 8.854\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\text{m}^{-2}.
  • Mass of electron me=9.109×10−31 kg=0.00055 u=0.511 MeV/c2m_e = 9.109\times10^{-31}\ \text{kg} = 0.00055\ \text{u} = 0.511\ \text{MeV}/c^2.
  • Mass of proton mp=1.00728 u=938.28 MeV/c2m_p = 1.00728\ \text{u} = 938.28\ \text{MeV}/c^2; mass of the hydrogen atom mH=1.007825 um_H = 1.007825\ \text{u}.
  • Mass of neutron mn=1.00866 u=939.57 MeV/c2m_n = 1.00866\ \text{u} = 939.57\ \text{MeV}/c^2.
  • 1 u=1.6605×10−27 kg1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}, and 1 u1\ \text{u} is equivalent to 931.5 MeV931.5\ \text{MeV}, i.e. 1 u=931.5 MeV/c21\ \text{u} = 931.5\ \text{MeV}/c^2.
  • Rydberg constant RH=1.097×107 m−1R_H = 1.097\times10^{7}\ \text{m}^{-1}; Bohr radius a0=0.053 nm=0.53 A˚a_0 = 0.053\ \text{nm} = 0.53\ \text{\text{Å}}.
  • Avogadro number NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\ \text{mol}^{-1} (textbook uses 6.02×10236.02\times10^{23}); 1 fm=10−15 m1\ \text{fm} = 10^{-15}\ \text{m}; 1 A˚=10−10 m1\ \text{\text{Å}} = 10^{-10}\ \text{m}.
  • 1 Bq=11\ \text{Bq} = 1 disintegration per second; 1 Ci=3.7×1010 Bq1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}.
  • c=3×108 m s−1c = 3\times10^{8}\ \text{m s}^{-1}; ln⁡2=0.693\ln 2 = 0.693; ln⁡x=2.303log⁡10x\ln x = 2.303\log_{10}x.

Geiger-Marsden experiment (distance of closest approach)

FormulaSymbols, units and conditions
K=14πε0(2e)(Ze)r1+r2K = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r_1+r_2}
  • Kinetic energy an alpha particle must start with to just reach the surface of a target nucleus of atomic number ZZ: at the stopping point all its kinetic energy has become electrostatic potential energy. r1r_1 and r2r_2 are the radii of the target nucleus and of the alpha particle, so r1+r2r_1+r_2 is the centre-to-centre separation. Answer in joule; divide by 1.6×10−131.6\times10^{-13} to get MeV.

Atomic spectra

FormulaSymbols, units and conditions
1λ=R(1n2−1m2)\dfrac{1}{\lambda} = R\left(\dfrac{1}{n^2} - \dfrac{1}{m^2}\right)
  • The empirical relation obeyed by every emission line of hydrogen. λ\lambda is the wavelength of the line, RR the Rydberg constant, nn fixes the series (n=1,2,3,4,5n = 1, 2, 3, 4, 5 for Lyman, Balmer, Paschen, Brackett, Pfund) and mm takes every integer value greater than nn. λ\lambda decreases as mm increases.
λlimit=n2R\lambda_{\text{limit}} = \dfrac{n^2}{R}
  • Series limit (shortest wavelength of a series), obtained by putting m→∞m \to \infty. Successive lines of a series crowd together towards this value.

Bohr's atomic model

FormulaSymbols, units and conditions
mevn2rn=14πε0Ze2rn2\dfrac{m_e v_n^2}{r_n} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r_n^2}
  • First postulate in equation form: the Coulomb attraction of the nucleus supplies the centripetal force for the circular orbit. ZZ is the atomic number (Z=1Z=1 for hydrogen).
mevnrn=nh2πm_e v_n r_n = \dfrac{nh}{2\pi}
  • Second postulate: the angular momentum of the electron in a stable orbit is an integral multiple of h2π\dfrac{h}{2\pi}. n=1,2,3,…n = 1, 2, 3,\ldots is the principal quantum number. Minimum angular momentum (for n=1n=1) is h2π\dfrac{h}{2\pi}.
Em−En=hνE_m - E_n = h\nu
  • Third postulate: a transition from a higher orbit mm to a lower orbit nn emits one photon of frequency ν\nu.
rn=(ε0h2πmeZe2)n2=a0n2(Z=1)r_n = \left(\dfrac{\varepsilon_0 h^2}{\pi m_e Z e^2}\right)n^2 = a_0 n^2 \quad (Z=1)
  • Radius of the nthn^{th} Bohr orbit; rn∝n2r_n \propto n^2 and rn∝1Zr_n \propto \dfrac{1}{Z}. For hydrogen a0=r1=0.053 nma_0 = r_1 = 0.053\ \text{nm}, so rn=0.053 n2 nmr_n = 0.053\,n^2\ \text{nm}. Ratio form for numericals: r1r2=n12n22\dfrac{r_1}{r_2} = \dfrac{n_1^2}{n_2^2}.
vn=Ze22ε0nhv_n = \dfrac{Ze^2}{2\varepsilon_0 n h}
  • Speed of the electron in the nthn^{th} orbit; vn∝Znv_n \propto \dfrac{Z}{n}. For hydrogen v1=2.18×106 m s−1v_1 = 2.18\times10^{6}\ \text{m s}^{-1}, so vn=2.18×106n m s−1v_n = \dfrac{2.18\times10^{6}}{n}\ \text{m s}^{-1}.
K.E=Ze28πε0rn,P.E=−Ze24πε0rn,T.E=−Ze28πε0rnK.E = \dfrac{Ze^2}{8\pi\varepsilon_0 r_n},\qquad P.E = -\dfrac{Ze^2}{4\pi\varepsilon_0 r_n},\qquad T.E = -\dfrac{Ze^2}{8\pi\varepsilon_0 r_n}
  • Kinetic, potential and total energy of the orbiting electron. Note P.E=−2 (K.E)P.E = -2\,(K.E) and T.E=−(K.E)T.E = -(K.E).
En=−meZ2e48ε02h2⋅1n2E_n = -\dfrac{m_e Z^2 e^4}{8\varepsilon_0^2 h^2}\cdot\dfrac{1}{n^2}
  • Total energy of the electron in the nthn^{th} orbit, in joule; En∝−1n2E_n \propto -\dfrac{1}{n^2}. The negative sign means the electron is bound.
En=−13.6 Z2n2 eVE_n = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV}
  • The same result in electronvolt. For hydrogen E1=−13.6 eVE_1 = -13.6\ \text{eV}, E2=−3.4 eVE_2 = -3.4\ \text{eV}, E3=−1.51 eVE_3 = -1.51\ \text{eV}, E4=−0.85 eVE_4 = -0.85\ \text{eV}, E∞=0E_\infty = 0. Ratio form: E1E2=n22n12\dfrac{E_1}{E_2} = \dfrac{n_2^2}{n_1^2}.
B.E=−T.E=mee48ε02h2n2=13.6n2 eV(Z=1)B.E = -T.E = \dfrac{m_e e^4}{8\varepsilon_0^2 h^2 n^2} = \dfrac{13.6}{n^2}\ \text{eV}\quad (Z=1)
  • Binding energy of the electron in the nthn^{th} orbit: the minimum energy needed to make it free, so that B.E+T.E=0B.E + T.E = 0.
Eexcitation=En−E1,Eionisation=E∞−En=−EnE_{\text{excitation}} = E_n - E_1,\qquad E_{\text{ionisation}} = E_\infty - E_n = -E_n
  • Excitation energy of the nthn^{th} state is measured from the ground state; ionisation energy from the state the electron is actually in. Ionisation energy of hydrogen in the ground state =13.6 eV= 13.6\ \text{eV}; minimum excitation energy =10.2 eV= 10.2\ \text{eV}.
νˉ=1λ=Z2mee48ε02h3c(1n2−1m2)=RHZ2(1n2−1m2)\bar\nu = \dfrac{1}{\lambda} = \dfrac{Z^2 m_e e^4}{8\varepsilon_0^2 h^3 c}\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right) = R_H Z^2\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)
  • Rydberg's formula for the wave number νˉ\bar\nu of an emitted line, with RH=mee48ε02h3c=1.097×107 m−1R_H = \dfrac{m_e e^4}{8\varepsilon_0^2 h^3 c} = 1.097\times10^{7}\ \text{m}^{-1}. Put Z=1Z=1 for hydrogen.
ν=Em−Enh=cλ\nu = \dfrac{E_m - E_n}{h} = \dfrac{c}{\lambda}
  • Frequency of the emitted photon; wave number νˉ=1λ\bar\nu = \dfrac{1}{\lambda} in m−1\text{m}^{-1}, frequency ν\nu in Hz.
T=2πrnvn,νrev=vn2πrn,ω=vnrnT = \dfrac{2\pi r_n}{v_n},\qquad \nu_{\text{rev}} = \dfrac{v_n}{2\pi r_n},\qquad \omega = \dfrac{v_n}{r_n}
  • Period, frequency and angular speed of revolution of the electron in the nthn^{th} orbit. Number of revolutions made in a time tt is tT\dfrac{t}{T}. The angular speed is largest for n=1n=1: ωmax⁡=2.18×1060.53×10−10≈4.1×1016 rad s−1\omega_{\max} = \dfrac{2.18\times10^{6}}{0.53\times10^{-10}} \approx 4.1\times10^{16}\ \text{rad s}^{-1}.
Ln=nh2π,ΔL=h2π(n2−n1)L_n = \dfrac{nh}{2\pi},\qquad \Delta L = \dfrac{h}{2\pi}(n_2 - n_1)
  • Angular momentum of the electron and the change in it during a transition, in kg m2s−1\text{kg m}^2\text{s}^{-1}.
2πrn=nλn,λn=2πrnn=hmevn2\pi r_n = n\lambda_n,\qquad \lambda_n = \dfrac{2\pi r_n}{n} = \dfrac{h}{m_e v_n}
  • De Broglie's standing-wave condition: the circumference of the nthn^{th} orbit holds exactly nn de Broglie wavelengths. It reproduces Bohr's second postulate.

Atomic nucleus

FormulaSymbols, units and conditions
A=Z+NA = Z + N
  • Mass number = number of protons (atomic number ZZ) + number of neutrons NN. A nuclide is written ZAX {}^{A}_{Z}\text{X}.
RX=R0A1/3,R0=1.2×10−15 mR_X = R_0 A^{1/3},\qquad R_0 = 1.2\times10^{-15}\ \text{m}
  • Radius of a nucleus of mass number AA. Ratio form: R1R2=(A1A2)1/3\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}. Volume ∝A\propto A.
ρ=3m4πR03≈2.3×1017 kg m−3\rho = \dfrac{3m}{4\pi R_0^3} \approx 2.3\times10^{17}\ \text{kg m}^{-3}
  • Density of nuclear matter, where mm is the average mass of a nucleon. AA cancels, so the density is the same for all nuclei. Compare osmium, the densest element, at 2.2×104 kg m−32.2\times10^{4}\ \text{kg m}^{-3}.
ρ=3M4πRX3\rho = \dfrac{3M}{4\pi R_X^3}
  • Working form when the actual nuclear mass MM (in kg, so multiply a value in u by 1.66×10−271.66\times10^{-27}) and radius are known.

Nuclear binding energy

FormulaSymbols, units and conditions
ΔM=(Zmp+Nmn)−M\Delta M = \left(Z m_p + N m_n\right) - M
  • Mass defect: the amount by which the mass MM of the nucleus falls short of the sum of the masses of its free constituent nucleons.
EB=ΔM c2=(Zmp+Nmn−M)c2E_B = \Delta M\,c^2 = \left(Z m_p + N m_n - M\right)c^2
  • Binding energy using nuclear masses. With masses in u, multiply the mass defect by 931.5931.5 to get EBE_B in MeV.
EB=(ZmH+Nmn−ZAM)c2E_B = \left(Z m_H + N m_n - {}^{A}_{Z}M\right)c^2
  • The form to use when the data are atomic masses: mHm_H is the mass of the hydrogen atom and ZAM {}^{A}_{Z}M is the atomic mass of the element, the ZZ electron masses having been added and subtracted.
EBA\dfrac{E_B}{A}
  • Binding energy per nucleon: the average energy needed to pull one nucleon out. Higher value means a more stable nucleus. The curve peaks near A=56A = 56 (iron) and is flat for A=50A = 50 to 8080.

Radioactive decays

FormulaSymbols, units and conditions
ZAX⟶Z−2A−4Y+24He{}^{A}_{Z}\text{X} \longrightarrow {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\text{He}
  • Alpha decay: AA falls by 4 and ZZ by 2. Occurs in all nuclei with A>210A > 210.
Q=(mX−mY−mHe)c2Q = \left(m_X - m_Y - m_{He}\right)c^2
  • QQ-value of alpha decay, using atomic masses; it equals the total kinetic energy shared by the products. In MeV when the bracket is in u and multiplied by 931.5931.5.
ZAX⟶Z+1AY+e−+νˉ,n⟶p+e−+νˉ{}^{A}_{Z}\text{X} \longrightarrow {}^{A}_{Z+1}\text{Y} + e^- + \bar\nu,\qquad n \longrightarrow p + e^- + \bar\nu
  • Beta-minus decay: a neutron in the nucleus becomes a proton; ZZ increases by 1, NN decreases by 1, AA is unchanged.
ZAX⟶Z−1AY+e++ν,p⟶n+e++ν{}^{A}_{Z}\text{X} \longrightarrow {}^{A}_{Z-1}\text{Y} + e^+ + \nu,\qquad p \longrightarrow n + e^+ + \nu
  • Beta-plus decay: a proton becomes a neutron; ZZ decreases by 1, NN increases by 1, AA is unchanged. It cannot happen to a free proton, only inside a nucleus.
Q=(mX−mY−me)c2Q = \left(m_X - m_Y - m_e\right)c^2
  • QQ-value of beta decay (textbook form), the neutrino mass being neglected. QQ is the maximum kinetic energy the beta particle can carry, which happens when the neutrino carries none.
ZAX∗⟶ZAX+γ,Eγ=E2−E1{}^{A}_{Z}\text{X}^{*} \longrightarrow {}^{A}_{Z}\text{X} + \gamma,\qquad E_\gamma = E_2 - E_1
  • Gamma decay: the daughter is the same nuclide in a lower nuclear energy state; nuclear level spacings are keV to MeV, so the photon is a gamma ray.
nα=Aparent−Adaughter4,nβ=2nα−(Zparent−Zdaughter)n_\alpha = \dfrac{A_{\text{parent}} - A_{\text{daughter}}}{4},\qquad n_\beta = 2n_\alpha - \left(Z_{\text{parent}} - Z_{\text{daughter}}\right)
  • Counting the alpha and beta-minus particles emitted in a whole decay series: fix AA from the alphas first, then correct ZZ with the betas.

Law of radioactive decay

FormulaSymbols, units and conditions
dNdt=−λN(t)\dfrac{dN}{dt} = -\lambda N(t)
  • Decay law in differential form: the number decaying per unit time is proportional to the number of undecayed nuclei present. λ\lambda is the decay constant, in s−1\text{s}^{-1} (or per hour, per day, per year — keep tt in the same unit).
N(t)=N0e−λtN(t) = N_0 e^{-\lambda t}
  • Number of parent nuclei left at time tt, starting from N0N_0 at t=0t = 0.
N=N0(12)t/T1/2N = N_0\left(\dfrac{1}{2}\right)^{t/T_{1/2}}
  • Equivalent working form when the elapsed time is a whole number of half-lives.
A(t)=−dNdt=λN(t)=A0e−λt,A0=λN0A(t) = -\dfrac{dN}{dt} = \lambda N(t) = A_0 e^{-\lambda t},\qquad A_0 = \lambda N_0
  • Activity: number of disintegrations per unit time. SI unit becquerel; 1 Ci=3.7×1010 Bq1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}. Activity falls off with exactly the same exponential as NN.
T1/2=ln⁡2λ=0.693λT_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}
  • Half-life: the time in which the number of parent nuclei falls to half its value.
τ=1λ,T1/2=τln⁡2=0.693 τ\tau = \dfrac{1}{\lambda},\qquad T_{1/2} = \tau\ln 2 = 0.693\,\tau
  • Average (mean) life of a nucleus of the species, and its relation to half-life. The mean life is always the longer of the two.
N=mMmol×NAN = \dfrac{m}{M_{\text{mol}}}\times N_A
  • Number of nuclei in a sample of mass mm grams whose molar mass (numerically the mass number) is MmolM_{\text{mol}} g per mol. Needed before any activity can be worked out from a given mass.
t=1λln⁡ ⁣(A0A)=2.303λlog⁡10 ⁣(A0A)t = \dfrac{1}{\lambda}\ln\!\left(\dfrac{A_0}{A}\right) = \dfrac{2.303}{\lambda}\log_{10}\!\left(\dfrac{A_0}{A}\right)
  • Age of a sample from its present activity per gram AA and the living or original value A0A_0 — the radiocarbon-dating formula. Same relation with N0/NN_0/N in place of A0/AA_0/A.
λ=1t2−t1ln⁡ ⁣(A1A2)\lambda = \dfrac{1}{t_2-t_1}\ln\!\left(\dfrac{A_1}{A_2}\right)
  • Decay constant from two activity readings taken at times t1t_1 and t2t_2; obtained by dividing A1=A0e−λt1A_1 = A_0e^{-\lambda t_1} by A2=A0e−λt2A_2 = A_0e^{-\lambda t_2}.
λpNp=λdNd\lambda_p N_p = \lambda_d N_d
  • Condition for the number of daughter nuclei to stay constant when the daughter is itself radioactive (radioactive equilibrium): daughters are produced as fast as they decay.

Nuclear energy

FormulaSymbols, units and conditions
92235U+n⟶92236U⟶56140Ba+3694Kr+2n+Q{}^{235}_{92}\text{U} + n \longrightarrow {}^{236}_{92}\text{U} \longrightarrow {}^{140}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + 2n + Q
  • A representative fission. Other channels of 92236U {}^{236}_{92}\text{U} give 53137I+3997Y+2n {}^{137}_{53}\text{I} + {}^{97}_{39}\text{Y} + 2n and 51133Sb+4199Nb+4n {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4n. The average number of neutrons released per fission is 2.72.7.
Q=(ΔM)c2=[Mparent−Mproducts]c2Q = \left(\Delta M\right)c^2 = \left[M_{\text{parent}} - M_{\text{products}}\right]c^2
  • Energy released in any nuclear reaction. For the fission above, ΔM=mU−(mBa+mKr+2mn)\Delta M = m_U - \left(m_{Ba} + m_{Kr} + 2m_n\right); multiply by 931.5931.5 for MeV. Typical fission releases about 200 MeV200\ \text{MeV}.
4p⟶α+2e++neutrinos+26.7 MeV4p \longrightarrow \alpha + 2e^+ + \text{neutrinos} + 26.7\ \text{MeV}
  • The net proton-proton cycle at the centre of the Sun (temperature about 107 K10^{7}\ \text{K}), which proceeds through several steps.
Q=[4mp−(mα+2me)]c2Q = \left[4m_p - \left(m_\alpha + 2m_e\right)\right]c^2
  • QQ-value of that fusion reaction; ignoring the energy carried away by the neutrinos it works out to 24.70 MeV24.70\ \text{MeV}.
Number of fissions per day=P tEone fission,m=P tEone fission×ANA\text{Number of fissions per day} = \dfrac{P\,t}{E_{\text{one fission}}},\qquad m = \dfrac{P\,t}{E_{\text{one fission}}}\times\dfrac{A}{N_A}
  • Mass of fuel consumed to deliver a thermal power PP for a time tt; convert Eone fissionE_{\text{one fission}} from MeV to joule by multiplying by 1.6×10−131.6\times10^{-13}.
Mavg=∑fimiM_{\text{avg}} = \sum f_i m_i
  • Weighted average atomic mass of an element from its isotopic masses mim_i and fractional abundances fif_i (with ∑fi=1\sum f_i = 1); used for the magnesium-abundance type of problem.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.11 subtopics
  1. 1Introduction
  2. 2Thomson's Atomic Model
  3. 3Geiger-Marsden Experiment
  4. 4Rutherford's Atomic Model
  5. 5Atomic Spectra1 board question
  6. 6Bohr's Atomic Model34 board questions
  7. 7Atomic Nucleus
  8. 8Nuclear Binding Energy1 board question
  9. 9Radioactive Decays1 board question
  10. 10Law of Radioactive Decay6 board questions
  11. 11Nuclear Energy1 board question
The whole chapter in brief

15.1 Introduction

  • Leucippus and Democritus first proposed indivisible atoms; Dalton's theory said matter is made of indestructible particles, atoms of a given element are identical, and atoms combine to form new substances.
  • J. J. Thomson's discovery of the electron destroyed "indestructible". Thomson's model was improved by Rutherford, and Rutherford's by Bohr.
  • Two numbers that frame the chapter: the nucleus is about 10510^5 times smaller in radius than the atom, and it carries all the positive charge and 99.9%99.9\% of the mass.

15.2 Thomson's atomic model

  • Evidence: in an evacuated discharge tube with a voltage across two electrodes, the cathode emits rays that glow on striking the glass behind the anode; the rays are negatively charged particles — electrons.
  • Model (1903): the atom is a sphere of uniform positive charge in which electrons are embedded — the plum-pudding model.
  • Successes: the atom is electrically neutral (total positive charge = total negative charge); only electrons can be emitted, since the positive charge is spread through the whole solid sphere; it explains the formation of ions and ionic compounds.
  • Failure: the alpha-scattering experiment showed the positive charge is not spread out at all.

15.3 Geiger-Marsden experiment

  • Suggested by Rutherford, performed by Geiger and Marsden between 1908 and 1913. Alpha particles (helium nuclei, charge +2e+2e) from a radioactive source were collimated into a narrow beam and fired at a thin gold foil; scattered particles produced scintillations on a surrounding screen viewed through a movable microscope.
  • Observations: most alpha particles went straight through; a few were scattered through various angles; only about 0.14%0.14\% were scattered through more than 0.1∘0.1^\circ; about one in 80008000 was deflected through more than 90∘90^\circ; a very few came back through nearly 180∘180^\circ.
  • Numerical use: for a head-on approach, the initial kinetic energy equals 14πε02Ze2r1+r2\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{r_1+r_2} at the point where the alpha particle stops.

15.4 Rutherford's atomic model

  • Thomson's model cannot explain the results: a diffuse positive charge has a small volume density and could only produce small deflections.
  • Rutherford's reasoning: a particle turned back must have met something massive and highly positively charged; since only a very few turn back, that something must be very small. Experiment gave its size as about 10 fm10\ \text{fm}, roughly 10−510^{-5} times the size of the atom, i.e. about 10−1510^{-15} of its volume. He called it the nucleus.
  • The model: all the positive charge and 99.9%99.9\% of the mass sit in a central nucleus; electrons revolve around it in circular orbits like planets round the Sun; the space between is mostly empty, which is why most alpha particles pass undeflected.
  • It also explains why atoms emit electrons but not positive particles — the heavy nucleus is unaffected by ordinary forces on the atom.

Difficulties with Rutherford's model

  • An electron in a circular orbit is accelerated (its direction changes), and by Maxwell's equations an accelerated charge radiates continuously.
  • Radiating, it loses energy, its orbital radius shrinks continuously, and it should spiral into the nucleus — the atom would collapse.
  • As it spirals in, its speed and hence the frequency of the radiation would change continuously, giving a continuous spectrum.
  • None of this is observed: atoms are stable, they do not radiate on their own, and when excited they emit only certain fixed frequencies. It also could not explain atomic spectra.

15.5 Atomic spectra

  • A heated metallic object gives a continuous spectrum through a prism. Hydrogen gas heated in a glass tube gives a line spectrum — only a few discrete wavelengths, with nothing in between.
  • Four visible hydrogen lines: 410410, 434434, 486486 and 656 nm656\ \text{nm}. These lines are called emission lines.
  • Lines fall into series, named from shortest to longest wavelength: Lyman, Balmer, Paschen, Brackett, Pfund.
  • Within a series the separation of successive lines decreases as the wavelength gets shorter, until they crowd up to the series limit n2R\dfrac{n^2}{R} reached as m→∞m\to\infty.
  • Empirical relation: 1λ=R(1n2−1m2)\dfrac{1}{\lambda} = R\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right), with nn fixed for a series and m>nm > n.
  • Every element gives its own unique set of wavelengths, so an element can be identified from its spectrum.
SeriesnnmmRegionSeries limit
Lyman12, 3, 4, …Ultraviolet911.6 A˚911.6\ \text{\text{Å}}
Balmer23, 4, 5, …Visible3646 A˚3646\ \text{\text{Å}}
Paschen34, 5, 6, …Infrared8204 A˚8204\ \text{\text{Å}}
Brackett45, 6, 7, …Near infrared14585 A˚14585\ \text{\text{Å}}
Pfund56, 7, 8, …Far infrared22790 A˚22790\ \text{\text{Å}}

15.6 Bohr's atomic model

  • Bohr kept Rutherford's picture and added quantum ideas to make the orbits stable and non-radiating.
  • Postulate 1: the electron revolves round the nucleus in a fixed circular orbit with constant speed, the electrostatic attraction providing the centripetal force — mevn2rn=14πε0Ze2rn2\dfrac{m_ev_n^2}{r_n} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r_n^2}.
  • Postulate 2: only orbits whose angular momentum is an integral multiple of h2π\dfrac{h}{2\pi} are allowed — mevnrn=nh2πm_e v_n r_n = \dfrac{nh}{2\pi}. Electrons in these stable orbits do not radiate.
  • Postulate 3: an electron may jump to an orbit of lower energy, emitting one photon of energy equal to the difference — Em−En=hνE_m - E_n = h\nu.
  • Standard results: rn=(ε0h2πmeZe2)n2=a0n2r_n = \left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2 = a_0n^2 with a0=0.053 nma_0 = 0.053\ \text{nm}; vn=Ze22ε0nhv_n = \dfrac{Ze^2}{2\varepsilon_0nh}; En=−meZ2e48ε02h2n2=−13.6Z2n2 eVE_n = -\dfrac{m_eZ^2e^4}{8\varepsilon_0^2h^2n^2} = -\dfrac{13.6Z^2}{n^2}\ \text{eV}.
  • Energy-level facts for hydrogen: E1=−13.6E_1 = -13.6, E2=−3.4E_2 = -3.4, E3=−1.51E_3 = -1.51, E4=−0.85E_4 = -0.85, E5=−0.544 eVE_5 = -0.544\ \text{eV}; levels crowd together and reach 00 as n→∞n\to\infty.
  • Three definitions the board asks together — excitation energy: energy needed to lift the electron from the ground state to a given excited state (minimum 10.2 eV10.2\ \text{eV} for hydrogen). Ionisation energy: minimum energy needed to set the electron free from the state it is in (13.6 eV13.6\ \text{eV} from the ground state). Binding energy: the energy with which the electron is bound, equal in magnitude to its total energy, B.E=−T.EB.E = -T.E.
  • Rydberg's formula follows from postulate 3: νˉ=1λ=RHZ2(1n2−1m2)\bar\nu = \dfrac{1}{\lambda} = R_HZ^2\left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right), with RH=mee48ε02h3c=1.097×107 m−1R_H = \dfrac{m_ee^4}{8\varepsilon_0^2h^3c} = 1.097\times10^7\ \text{m}^{-1}.
  • Useful standard result: the frequency of the first Lyman line, RHc(1−14)=3RHc4R_Hc\left(1-\dfrac{1}{4}\right) = \dfrac{3R_Hc}{4}, equals the difference of the limiting frequencies of the Lyman (RHcR_Hc) and Balmer (RHc4\dfrac{R_Hc}{4}) series.

Limitations of Bohr's model

  • It cannot explain the spectra of atoms other than hydrogen, nor the fine multiple components of some hydrogen lines.
  • It gives no account of why emission lines differ in intensity.
  • It is theoretically unsatisfactory: the quantisation condition for stable orbits is simply assumed, with no basis.

De Broglie's explanation of the second postulate

  • Treat the orbiting electron as a standing wave rather than a particle. A standing wave fits only if the orbit length is a whole number of wavelengths: 2πrn=nλn2\pi r_n = n\lambda_n.
  • With λn=hpn\lambda_n = \dfrac{h}{p_n} this gives pn=nh2πrnp_n = \dfrac{nh}{2\pi r_n}, hence Ln=pnrn=nh2πL_n = p_nr_n = \dfrac{nh}{2\pi} — exactly Bohr's second postulate.
  • Diagram to describe: a circular orbit with a continuous wavy line drawn along it closing smoothly on itself; for n=4n=4, four complete wavelengths fit round the circle.

15.7 Atomic nucleus

  • The nucleus is made of protons and neutrons, together called nucleons. A proton carries +e+e; a neutron is neutral and slightly heavier. A proton is about 18361836 times as massive as an electron.
  • ZZ = number of protons = number of electrons in the neutral atom = atomic number; NN = number of neutrons; A=Z+NA = Z+N = mass number. Notation ZAX {}^{A}_{Z}\text{X}, e.g. 11H {}^{1}_{1}\text{H}, 612C {}^{12}_{6}\text{C}, 816O {}^{16}_{8}\text{O}.
  • Chemical properties are decided by ZZ alone.
TermSameDifferentExample
IsotopesNumber of protons ZZNumber of neutrons NN, mass number AA11H {}^{1}_{1}\text{H}, 12H {}^{2}_{1}\text{H}, 13H {}^{3}_{1}\text{H}
IsobarsMass number AAZZ and NN13H {}^{3}_{1}\text{H} and 23He {}^{3}_{2}\text{He}
IsotonesNumber of neutrons NNZZ and AA13H {}^{3}_{1}\text{H} and 24He {}^{4}_{2}\text{He}

Units of mass

  • Kilogram: me=9.109383×10−31m_e = 9.109383\times10^{-31}, mp=1.672623×10−27m_p = 1.672623\times10^{-27}, mn=1.674927×10−27 kgm_n = 1.674927\times10^{-27}\ \text{kg}.
  • Unified atomic mass unit: 1 u1\ \text{u} is one twelfth of the mass of a neutral carbon-12 atom in its lowest electronic state, =1.6605402×10−27 kg= 1.6605402\times10^{-27}\ \text{kg}. In this unit me=0.00055m_e = 0.00055, mp=1.00728m_p = 1.00728, mn=1.00866 um_n = 1.00866\ \text{u}, and the hydrogen atom is 1.007825 u1.007825\ \text{u}.
  • Energy equivalent: E=mc2E = mc^2, and masses are quoted in MeV/c2\text{MeV}/c^2 with 1 u=931.5 MeV/c21\ \text{u} = 931.5\ \text{MeV}/c^2. So me=0.511m_e = 0.511, mp=938.28m_p = 938.28, mn=939.57 MeV/c2m_n = 939.57\ \text{MeV}/c^2.

Size and density of nuclei

  • RX=R0A1/3R_X = R_0A^{1/3} with R0=1.2×10−15 mR_0 = 1.2\times10^{-15}\ \text{m}, so the nuclear volume is proportional to AA.
  • ρ=3m4πR03=2.3×1017 kg m−3\rho = \dfrac{3m}{4\pi R_0^3} = 2.3\times10^{17}\ \text{kg m}^{-3}: independent of AA, the same for every nucleus, and thirteen orders of magnitude greater than osmium, the densest element at 2.2×104 kg m−32.2\times10^{4}\ \text{kg m}^{-3}.

Nuclear forces

  • The strong (nuclear) force acts between nucleons and is mostly attractive; it is what holds the nucleus together against the electrostatic repulsion of the protons.
  • Three properties the board wants: (i) it is the strongest of the subatomic forces, some 5050 to 6060 times stronger than the electrostatic force; (ii) its range is only a few fm, unlike the infinite range of electrostatic and gravitational forces, and it is negligible beyond that; (iii) it is charge independent — the same between nn–nn, pp–pp and nn–pp at the same separation.

15.8 Nuclear binding energy

  • Mass defect: the difference between the actual mass of a nucleus and the sum of the masses of its constituent nucleons, ΔM=Zmp+Nmn−M\Delta M = Zm_p + Nm_n - M. The nucleus is always lighter than its parts.
  • Binding energy: the energy that must be supplied to separate all the nucleons and take them far apart, EB=ΔMc2E_B = \Delta M c^2; the same energy is released when the nucleus is assembled from free nucleons.
  • With atomic-mass data use EB=(ZmH+Nmn−ZAM)c2E_B = \left(Zm_H + Nm_n - {}^{A}_{Z}M\right)c^2 — the ZZ electron masses added and subtracted cancel.
  • Binding energy per nucleon EBA\dfrac{E_B}{A} is the average energy needed to free one nucleon and is the measure of stability: the larger it is, the more stable the nucleus.
  • Shape of the EBA\dfrac{E_B}{A} against AA curve, to describe in words: it starts very low, deuterium having the smallest value and being the least stable nucleus; it rises steeply, flattens into a plateau for AA between 5050 and 8080 where the nuclei are most stable, peaks near A=56A = 56 (iron), and then falls gradually for A>80A > 80.
  • The binding energy of the hydrogen nucleus, a single proton, is zero.
  • Worked values worth remembering: alpha particle 28.29 MeV28.29\ \text{MeV} (7.07 MeV7.07\ \text{MeV} per nucleon), 37Li {}^{7}_{3}\text{Li} about 39.2 MeV39.2\ \text{MeV}, americium-244 about 7.5 MeV7.5\ \text{MeV} per nucleon.

15.9 Radioactive decays

  • Unstable nuclei change into other nuclei by emitting particles. The decaying nucleus is the parent, the one produced is the daughter, and the process is radioactive decay, discovered by Becquerel; the term radioactivity was coined by Marie Curie.
DecayEmittedChange in AAChange in ZZQQ-value
Alpha24He {}^{4}_{2}\text{He} nucleus−4-4−2-2(mX−mY−mHe)c2\left(m_X-m_Y-m_{He}\right)c^2
Beta minuselectron and antineutrino00+1+1(mX−mY−me)c2\left(m_X-m_Y-m_e\right)c^2
Beta pluspositron and neutrino00−1-1(mX−mY−me)c2\left(m_X-m_Y-m_e\right)c^2
Gammahigh-energy photon0000Eγ=E2−E1E_\gamma = E_2-E_1
  • Alpha decay happens in all nuclei with A>210A > 210: too many protons, electrostatic repulsion beats the nuclear attraction, so the nucleus sheds protons two at a time. Example: 83212Bi⟶81208Tl+α {}^{212}_{83}\text{Bi} \longrightarrow {}^{208}_{81}\text{Tl} + \alpha.
  • Beta-minus decay converts a neutron into a proton, n→p+e−+νˉn \to p + e^- + \bar\nu. Example: 2760Co⟶2860Ni+e−+νˉ {}^{60}_{27}\text{Co} \longrightarrow {}^{60}_{28}\text{Ni} + e^- + \bar\nu.
  • Beta-plus decay converts a proton into a neutron, p→n+e++νp \to n + e^+ + \nu. Since the neutron is heavier, this cannot happen to a free proton; inside a nucleus the extra energy comes from the rest of the nucleus. Example: 1122Na⟶1022Ne+e++ν {}^{22}_{11}\text{Na} \longrightarrow {}^{22}_{10}\text{Ne} + e^+ + \nu.
  • Gamma decay follows an alpha or beta decay that leaves the daughter in an excited nuclear state. Example: 57Co {}^{57}\text{Co} beta-plus decays to 57Fe {}^{57}\text{Fe} in a state 136 keV136\ \text{keV} above the ground state, which de-excites either by one 136 keV136\ \text{keV} photon or by a 122 keV122\ \text{keV} photon followed by a 14 keV14\ \text{keV} photon.
  • Atomic energy levels are separated by a few eV, nuclear levels by keV to MeV — hence atoms radiate from ultraviolet to radio, nuclei radiate gamma rays.
  • The daughter is often unstable too, giving a decay series; 238U {}^{238}\text{U} takes fourteen alpha and beta decays to reach stable 206Pb {}^{206}\text{Pb}.
  • The QQ-value equals the total kinetic energy of the products, so it is also the maximum kinetic energy of the beta particle (the case where the neutrino gets nothing).

15.10 Law of radioactive decay

  • Statement: the number of nuclei undergoing decay per unit time is proportional to the number of undecayed nuclei present at that instant.
  • dN=−λN(t)dtdN = -\lambda N(t)dt, integrating to N(t)=N0e−λtN(t) = N_0e^{-\lambda t}; λ\lambda is the decay constant and the minus sign says NN falls with time.
  • Activity A(t)=−dNdt=λN(t)=A0e−λtA(t) = -\dfrac{dN}{dt} = \lambda N(t) = A_0e^{-\lambda t}, measured in becquerel (11 decay per second) or curie (1 Ci=3.7×1010 Bq1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}).
  • Half-life T1/2=0.693λT_{1/2} = \dfrac{0.693}{\lambda}: putting N=N02N = \dfrac{N_0}{2} in the decay law. After each further half-life the number halves again — N0→N02→N04→N08N_0 \to \dfrac{N_0}{2} \to \dfrac{N_0}{4} \to \dfrac{N_0}{8} — and never reaches zero.
  • Average (mean) life τ=1λ\tau = \dfrac{1}{\lambda}, found by averaging tt over λN0e−λtdt\lambda N_0e^{-\lambda t}dt; hence T1/2=0.693 τT_{1/2} = 0.693\,\tau, so τ>T1/2\tau > T_{1/2}.
  • Which nucleus decays and when is unpredictable; only the statistical behaviour of a large number is described by the law.
  • Other exponential decays for comparison: amplitude of a damped pendulum A=A0e−btA = A_0e^{-bt}; Newton cooling T=T0e−ktT = T_0e^{-kt}; discharge of a capacitor Q=Q0e−t/RCQ = Q_0e^{-t/RC}.
QuantityHalf-life T1/2T_{1/2}Average life τ\tau
MeaningTime for NN to fall to N02\dfrac{N_0}{2}Mean lifetime of one nucleus
Relation to λ\lambda0.693λ\dfrac{0.693}{\lambda}1λ\dfrac{1}{\lambda}
Relative sizeSmallerLarger, τ=1.44 T1/2\tau = 1.44\,T_{1/2}

15.11 Nuclear energy

  • Chemical reactions release a few eV per reaction, nuclear reactions a few MeV — about a million times more energy per unit mass of fuel.
  • Both routes to nuclear energy exploit the same fact: the products have a higher binding energy per nucleon than the reactants, so they are lighter, and the missing mass appears as energy, Q=(ΔM)c2Q = (\Delta M)c^2.

Nuclear fission

  • Splitting a heavy nucleus into two lighter nuclei with release of energy. Middle-weight nuclei (AA from 5050 to 8080) have the highest EBA\dfrac{E_B}{A}, so a nucleus of A≈230A \approx 230 breaking into two fragments of AA between 5050 and 150150 releases energy.
  • 92236U {}^{236}_{92}\text{U} is the working fuel nucleus. It is not found in nature; it is made by letting naturally occurring 92235U {}^{235}_{92}\text{U} absorb a slow neutron. Natural uranium is more than 99%99\% 92238U {}^{238}_{92}\text{U} and less than 1%1\% 92235U {}^{235}_{92}\text{U}.
  • Three textbook fission channels of 92236U {}^{236}_{92}\text{U}: to 53137I+3997Y+2n {}^{137}_{53}\text{I} + {}^{97}_{39}\text{Y} + 2n; to 56140Ba+3694Kr+2n {}^{140}_{56}\text{Ba} + {}^{94}_{36}\text{Kr} + 2n; to 51133Sb+4199Nb+4n {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4n. About 177177 to 200 MeV200\ \text{MeV} is released, appearing as kinetic energy of the products, i.e. as heat.
  • Chain reaction: each fission releases on average 2.72.7 neutrons, which are absorbed by further 235U {}^{235}\text{U} nuclei; unchecked, the number of fissions cascades and the result is an explosion.
  • Nuclear reactor: a device in which fission is carried out in a controlled way, the chain reaction being prevented from running away; the heat produced is carried off and converted to electricity with turbines. India has 22 reactors, the largest at Kudankulam in Tamil Nadu; the USA generates the most nuclear power.

Nuclear fusion

  • Two light nuclei (A<40A < 40) fuse into a heavier one with higher EBA\dfrac{E_B}{A}; the product is lighter than the reactants and MeV of energy is released.
  • Conditions: the nuclei must come within about 1 fm1\ \text{fm} so the nuclear force can act. Atoms must first be stripped of electrons, and the bare nuclei must be given enough kinetic energy to overcome their mutual electrostatic repulsion — hence the need for very high temperature, which rises further with ZZ.
  • In the Sun, at about 107 K10^{7}\ \text{K}, four protons fuse in several steps: 4p→α+2e++neutrinos+26.7 MeV4p \to \alpha + 2e^+ + \text{neutrinos} + 26.7\ \text{MeV}; ignoring neutrino energy, Q=[4mp−(mα+2me)]c2=24.70 MeVQ = \left[4m_p - (m_\alpha + 2m_e)\right]c^2 = 24.70\ \text{MeV}.
  • Fusion is the energy source of stars and has been running in the Sun for about 4.54.5 billion years. Stellar fusion stops at iron, because iron has the highest EBA\dfrac{E_B}{A} and fusing past it would need energy supplied; heavier elements are made in stellar explosions.
  • Deuterium, helium, lithium, beryllium and boron were not made in stars but within the first 200200 seconds after the big bang.
FeatureNuclear fissionNuclear fusion
ProcessHeavy nucleus splits into two lighter nucleiTwo light nuclei combine into a heavier one
Nuclei involvedAA large, around 230230AA small, below about 4040
TriggerBombardment by a slow neutronVery high temperature and pressure
ProductsTwo fragments plus 2 to 4 neutronsOne heavier nucleus plus light particles
Energy per reactionAbout 200 MeV200\ \text{MeV}About 26.7 MeV26.7\ \text{MeV} for the solar cycle
Where usedNuclear reactors, atom bombsStars, the Sun, hydrogen bombs
FeatureNuclear reactorNuclear bomb
Chain reactionControlledUncontrolled
Rate of energy releaseSlow and steadyEnormous energy in a very short time
PurposeGenerate electricityDestruction

Harmful effects

  • An uncontrolled chain reaction in nuclear fuel releases an enormous amount of energy in a very short time; this is the principle of the atom bomb, which uses fission alone or fission plus fusion. Two such bombs destroyed Hiroshima and Nagasaki in August 1945.

Traps that cost marks

  • Angular momentum is quantised, not energy directly: write mevnrn=nh2πm_ev_nr_n = \dfrac{nh}{2\pi}, not mevnrn=nhm_ev_nr_n = nh.
  • rn∝n2r_n \propto n^2 but rn∝1Zr_n \propto \dfrac{1}{Z} — so the first orbit is smallest for singly ionised helium (Z=2Z=2), not for hydrogen, deuterium or tritium, which all have Z=1Z=1. The fourth orbit is smaller than the eighth by a factor of 44, not 22.
  • En=−13.6n2 eVE_n = -\dfrac{13.6}{n^2}\ \text{eV} carries a minus sign, and energy increases (becomes less negative) with nn. Dropping the sign in an excitation- or ionisation-energy question loses the mark.
  • Excitation energy is measured from the ground state; ionisation energy from the state the electron is currently in. For an electron already in n=5n=5, the ionisation energy is 0.544 eV0.544\ \text{eV}, not 13.6 eV13.6\ \text{eV}.
  • Longest wavelength in a series is m=n+1m = n+1; shortest (the series limit) is m→∞m \to \infty. Small energy gaps mean long wavelengths, so the n=8n=8 to n=7n=7 transition gives a longer wavelength than n=2n=2 to n=1n=1.
  • Balmer is the visible series; Lyman is ultraviolet. Do not put Balmer in the UV.
  • Nuclear radius depends on AA as A1/3A^{1/3}, and volume and mass on AA — but density does not depend on AA at all. That is the standard MCQ.
  • Choose the binding-energy formula to match the data: nuclear masses go with mpm_p and mnm_n; atomic masses go with mHm_H and mnm_n. Using mHm_H with a nuclear mass double-counts ZZ electrons.
  • Mass defect is a mass (in u or kg); binding energy is an energy. Multiply by 931.5931.5 only when the mass is in u, and the answer is then in MeV.
  • After two half-lives the number left is N4\dfrac{N}{4}, not N2\dfrac{N}{2} — a half-life is not "half the total life".
  • Average life is longer than half-life: τ=T1/20.693=1.44 T1/2\tau = \dfrac{T_{1/2}}{0.693} = 1.44\,T_{1/2}. Do not invert the relation.
  • Keep λ\lambda and tt in the same time unit throughout a decay problem, and convert to s−1\text{s}^{-1} only when the answer is wanted in becquerel.
  • Activity questions that give a mass need N=mMmolNAN = \dfrac{m}{M_{\text{mol}}}N_A first; the activity is λN\lambda N, never λm\lambda m.
  • In beta decay the mass number does not change; only ZZ does. Balance both AA and ZZ on the two sides of every nuclear equation, and remember the neutrino or antineutrino.
  • In fission the neutrons on the product side carry mass — include 2mn2m_n (or 4mn4m_n) in the QQ-value calculation.
  • Fusion needs high temperature to beat electrostatic repulsion, not to break bonds; fission needs a slow (thermal) neutron, not a fast one.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 3 have come up in two or more papers.5
  1. 1
    Radius of the nthn^{th} Bohr orbit Repeats
    Derive an expression for the radius of the nthn^{th} Bohr orbit of the electron in a hydrogen atom.

    Asked in: March 2015; March 2022 (Q.6, 2 marks); February 2026 (Q.6, 2 marks)

    Show the derivation
    Expression for radius of Bohr orbit in atom:
    Step 1
    Let, mem_e = mass of electron, −e-e = charge on electron, rnr_n = radius of nthn^{th} Bohr's orbit, +e+e = charge on nucleus, vnv_n = linear velocity of electron in nthn^{th} orbit, Z = number of electrons in an atom, n = principal quantum number.
    Step 2
    From Bohr's first postulate,
    Coulomb's force FeF_e = Centripetal force FcpF_{cp}
    ∴Ze24πε0rn2=mevn2rn\therefore\dfrac{Ze^2}{4\pi\varepsilon_0r_n^2}=\dfrac{m_ev_n^2}{r_n}
    ∴vn2=Ze24πε0rnme....(1)\therefore v_n^2=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}\qquad....(1)
    Step 3
    According to Bohr's second postulate,
    mernvn=nh2πm_er_nv_n=\dfrac{nh}{2\pi}
    ∴me2vn2rn2=n2h24π2\therefore m_e^2v_n^2r_n^2=\dfrac{n^2h^2}{4\pi^2}
    ∴vn2=n2h24π2me2rn2....(2)\therefore v_n^2=\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}\qquad....(2)
    Step 4
    From equations (1) and (2),
    n2h24π2me2rn2=Ze24πε0rnme\dfrac{n^2h^2}{4\pi^2m_e^2r_n^2}=\dfrac{Ze^2}{4\pi\varepsilon_0r_nm_e}
    ∴rn=n2h2ε0πmeZe2\therefore r_n=\dfrac{n^2h^2\varepsilon_0}{\pi m_eZe^2}
    ∴rn=(ε0h2πmeZe2)n2=a0n2....(3)\therefore r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2=a_0n^2\qquad....(3)
    This is the required expression for radius of nthn^{th} orbit.
  2. 2
    Half-life of a radioactive material from the law of radioactive decay
    Derive an expression for 'Half Life Time' of a radioactive material using the 'Law of Radioactive Decay'.

    Asked in: July 2023 (Q.11, 2 marks)

    Show the derivation

    Expression for half life period: From law of radioactive decay, N(t)=N0e−λtN(t)=N_0e^{-\lambda t} At t=T1/2t=T_{1/2}, N(t)=N02N(t)=\dfrac{N_0}{2} ∴N02=N0e−λT(1/2)\therefore\dfrac{N_0}{2}=N_0e^{-\lambda T_{(1/2)}} ∴eλT(1/2)=2\therefore e^{\lambda T_{(1/2)}}=2 ∴λT1/2=log⁡e2=0.693\therefore\lambda T_{1/2}=\log_e2=0.693 ∴T1/2=0.693λ\therefore T_{1/2}=\dfrac{0.693}{\lambda} This is the required expression for half life period of radioactive substance.

  3. 3
    Law of radioactive decay; SI unit of activity Repeats
    Derive the decay law of radioactivity and state the SI unit of 'activity'.

    Asked in: March 2013; June 2025 (Q.21, 3 marks)

    Show the derivation
    StatementThe number of nuclei undergoing the decay per unit time is proportional to the number of unchanged nuclei present at that instant.
    Derivation:
    If 'N(t)' is the number of parent nuclei present at any instant 't', 'dN' is the number of nuclei disintegrated in short interval of time 'dt', then,
    dN∝−N(t)dtdN\propto-N(t)dt
    dN=−λN(t)dtdN=-\lambda N(t)dt
    where, λ\lambda is known as decay constant or disintegration constant.
    The negative sign indicates disintegration of atoms.
    Integrating both sides of equation,
    ∫N0N(t)dNN(t)=∫0t−λ dt\int_{N_0}^{N(t)}\dfrac{dN}{N(t)}=\int_0^t-\lambda\,dt
    where, N0N_0 is number of parent atoms at time t = 0.
    ∴log⁡eN(t)N0=−λt\therefore\log_e\dfrac{N(t)}{N_0}=-\lambda t
    ∴N(t)=N0e−λt\therefore N(t)=N_0e^{-\lambda t}
    This is the decay law of radioactivity.
    Refer subtopic 15.10: Q. No. 1
  4. 4
    Wavelength of spectral lines in a hydrogen atom (from the electron energy expression)
    What is ionization energy? Assuming the expression for energy of an electron, derive an expression for the wavelength of spectral lines in a hydrogen atom.

    Asked in: July 2024 (Q.20, 3 marks)

    Show the derivation
    Step 1
    Let, EmE_m = Energy of electron in mthm^{th} higher orbit, EnE_n = Energy of electron in nthn^{th} lower orbit.
    Step 2
    According to Bohr's third postulate,
    Em−En=hνE_m-E_n=h\nu
    ∴ν=Em−Enh....(1)\therefore\nu=\dfrac{E_m-E_n}{h}\qquad....(1)
    Step 3
    But Em=−Z2mee48ε02h2m2....(2)E_m=-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2m^2}\qquad....(2)
    En=−Z2mee48ε02h2n2....(3)E_n=-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2n^2}\qquad....(3)
    Step 4
    From equations (1), (2) and (3),
    ν=−Z2mee48ε02h2m2−(−Z2mee48ε02h2n2)h\nu=\dfrac{-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2m^2}-\left(-\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^2n^2}\right)}{h}
    ∴ν=Z2mee48ε02h3[−1m2+1n2]\therefore\nu=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3}\left[-\dfrac{1}{m^2}+\dfrac{1}{n^2}\right]
    ∴cλ=Z2mee48ε02h3[1n2−1m2][∵ν=cλ]\therefore\dfrac{c}{\lambda}=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3}\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]\qquad\left[\because\nu=\dfrac{c}{\lambda}\right]
    where, c = speed of electromagnetic radiation
    ∴1λ=Z2mee48ε02h3c[1n2−1m2]\therefore\dfrac{1}{\lambda}=\dfrac{Z^2m_ee^4}{8\varepsilon_0^2h^3c}\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]
    Step 5
    But, mee48ε02h3c=RH\dfrac{m_ee^4}{8\varepsilon_0^2h^3c}=R_H = Rydberg's constant
    =1.097×107 m−1=1.097\times10^7\text{ m}^{-1}
    ∴1λ=RHZ2[1n2−1m2]....(4)\therefore\dfrac{1}{\lambda}=R_HZ^2\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]\qquad....(4)
    Equation (4) represents Rydberg's formula for atomic spectrum.
    Step 6
    1λ\dfrac{1}{\lambda} is called wave number (νˉ\bar\nu) of the line.
    ∴νˉ=1λ=RHZ2[1n2−1m2]\therefore\bar\nu=\dfrac{1}{\lambda}=R_HZ^2\left[\dfrac{1}{n^2}-\dfrac{1}{m^2}\right]
  5. 5
    Energy of an electron in the nth Bohr orbit and its binding energy Repeats
    Derive an expression for the total energy of an electron in the nth Bohr orbit. Hence show that the energy of the electron is inversely proportional to the square of the principal quantum number. Also define binding energy.

    Asked in: October 2014; July 2018

    Show the derivation
    Expression for total energy of electron:
    Step 1
    Kinetic energy:
    Let, mem_e = mass of electron, rnr_n = radius of nthn^{th} orbit of Bohr's hydrogen atom, vnv_n = velocity of electron, −e-e = charge of electron, +e+e = charge on the nucleus, Z = number of electrons in an atom.
    According to Bohr's first postulate,
    mevn2rn=14πε0×Ze2rn2\dfrac{m_ev_n^2}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{Ze^2}{r_n^2}
    where, ε0\varepsilon_0 is permittivity of free space.
    ∴mevn2=Z4πε0×e2rn....(1)\therefore m_ev_n^2=\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\qquad....(1)
    The revolving electron in the circular orbit has linear speed and hence it possesses kinetic energy.
    It is given by, K.E=12mevn2K.E=\dfrac{1}{2}m_ev_n^2
    ∴K.E=12×(Z4πε0×e2rn)....[From equation (1)]\therefore K.E=\dfrac{1}{2}\times\left(\dfrac{Z}{4\pi\varepsilon_0}\times\dfrac{e^2}{r_n}\right)\qquad....\text{[From equation (1)]}
    ∴K.E=Ze28πε0rn....(2)\therefore K.E=\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(2)
    Step 2
    Potential energy:
    Potential energy of electron is given by, P.E =V(−e)=V(-e)
    where, V = electric potential at any point due to charge on nucleus, −e-e = charge on electron.
    In this case,
    ∴P.E=14πε0×ern×(−Ze)\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{e}{r_n}\times(-Ze)
    ∴P.E=14πε0×−Ze2rn\therefore P.E=\dfrac{1}{4\pi\varepsilon_0}\times\dfrac{-Ze^2}{r_n}
    ∴P.E=−Ze24πε0rn....(3)\therefore P.E=-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\qquad....(3)
    Step 3
    Total energy:
    The total energy of the electron in any orbit is its sum of P.E and K.E.
    ∴T.E=K.E+P.E\therefore T.E=K.E+P.E
    =(Ze28πε0rn)+(−Ze24πε0rn)....[From equations (2) and (3)]=\left(\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\right)+\left(-\dfrac{Ze^2}{4\pi\varepsilon_0r_n}\right)\qquad....\text{[From equations (2) and (3)]}
    ∴T.E=−Ze28πε0rn....(4)\therefore T.E=-\dfrac{Ze^2}{8\pi\varepsilon_0r_n}\qquad....(4)
    Step 4
    But, rn=(ε0h2πmeZe2)×n2r_n=\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)\times n^2
    Substituting for rnr_n in equation (4),
    ∴T.E=−18πε0×Ze2(ε0h2πmeZe2)n2=−18πε0×Z2e2πmee2ε0h2n2\therefore T.E=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Ze^2}{\left(\dfrac{\varepsilon_0h^2}{\pi m_eZe^2}\right)n^2}=-\dfrac{1}{8\pi\varepsilon_0}\times\dfrac{Z^2e^2\pi m_ee^2}{\varepsilon_0h^2n^2}
    ∴T.E=−(meZ2e48ε02h2)×1n2....(5)\therefore T.E=-\left(\dfrac{m_eZ^2e^4}{8\varepsilon_0^2h^2}\right)\times\dfrac{1}{n^2}\qquad....(5)
    This is required expression for energy of electron in nthn^{th} orbit of Bohr's hydrogen atom.
    Step 5
    The negative sign in equation (5) shows that the electron is bound to the nucleus by an attractive force and hence energy must be supplied to the electron in order to make it free from the influence of the nucleus.
    Step 6
    As m, e, ε0\varepsilon_0, and h in equation (5) are constant,
    ∴me48ε02h2=constant\therefore\dfrac{me^4}{8\varepsilon_0^2h^2}=\text{constant}
    Using equation (5),
    T.E=constant×1n2T.E=\text{constant}\times\dfrac{1}{n^2}
    ∴T.E∝1n2\therefore T.E\propto\dfrac{1}{n^2}
    Binding energy: Binding energy of an electron is the minimum energy required to make it free from the nucleus.

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