Atoms and Nuclei — Easy Practice (Set 1) — JEE Main 2027 Physics Notes

Quick Recap — The Bohr Atom

Rutherford: alpha-scattering revealed a tiny, massive, positively charged nucleus; the atom is mostly empty space.
Bohr's postulates: electrons occupy stable quantized orbits with angular momentum $L = \dfrac{nh}{2\pi}$ , and radiate only when jumping between orbits.
Energy levels (hydrogen): $E_n = -\dfrac{13.6}{n^2}$ eV; the ground state is $-13.6$ eV and the ionization energy is $13.6$ eV.
Bohr radius: $r_n = 0.53\,n^2$ Å (so $r_n\propto n^2$ ).
Spectrum (Rydberg): $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ ; Lyman (UV, to $n=1$ ), Balmer (visible, to $n=2$ ), Paschen (IR, to $n=3$ ).

Beyond-NCERT JEE Formulae

The practice sets state the Bohr, spectral and nuclear results one line at a time. Here they are gathered into a single working sheet, with the $Z$ -scalings, the exact constants, the shortcut forms JEE rewards, and the traps that quietly cost marks. Fix the hydrogen anchors and everything else is one scaling away.

1. Bohr model — hydrogen-like atoms (one electron, nuclear charge $+Ze$ )

Radius: $r_n=0.529\dfrac{n^2}{Z}$ angstrom, so $r_n\propto\dfrac{n^2}{Z}$ .
Energy: $E_n=-13.6\dfrac{Z^2}{n^2}$ eV (bound, hence negative), so $E_n\propto\dfrac{Z^2}{n^2}$ .
Speed: $v_n=2.18\times10^6\,\dfrac{Z}{n}$ m/s, i.e. $\dfrac{1}{137}$ of light-speed times $\dfrac{Z}{n}$ .
Energy split: $KE=-E_n=+13.6\dfrac{Z^2}{n^2}$ eV, potential energy $PE=2E_n$ , total $=E_n$ ; hence $PE=-2\,KE$ .
Period and frequency of revolution: $T_n\propto\dfrac{n^3}{Z^2}$ and $f_n\propto\dfrac{Z^2}{n^3}$ .
de Broglie condition: $2\pi r_n=n\lambda$ — the orbit holds exactly $n$ electron wavelengths.
Distinct emission lines from level $n$ : a gas fully excited to level $n$ can radiate $\dfrac{n(n-1)}{2}$ wavelengths.
When to use: any single-electron species (H, He $^+$ , Li $^{2+}$ ); read off $Z$ and $n$ and scale from the hydrogen values.
[JEE Tip] Almost every Bohr quantity is a power of $\dfrac{Z}{n}$ : energy and $KE$ as $\left(\dfrac{Z}{n}\right)^2$ , speed as $\dfrac{Z}{n}$ , radius as $\dfrac{n^2}{Z}$ . Memorise the three anchors ( $0.529$ angstrom, $-13.6$ eV, $2.18\times10^6$ m/s) and you never re-derive.

2. Rydberg formula and spectral series

Wavenumber: $\dfrac1\lambda=RZ^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ with $n_2>n_1$ and $R=1.097\times10^7$ per metre.
Energy-to-wavelength shortcut: $\lambda(\text{in nm})=\dfrac{1240}{\Delta E(\text{in eV})}$ , where $\Delta E=E_{n_2}-E_{n_1}$ .
Series, named by the landing level $n_1$ : Lyman $n_1=1$ (UV), Balmer $n_1=2$ (visible), Paschen $n_1=3$ (infrared).
Series limit, the shortest wavelength as $n_2\to\infty$ : $\lambda_{\min}=\dfrac{n_1^2}{RZ^2}$ .
Longest line of a series, the smallest gap at $n_2=n_1+1$ : the reddest, least energetic line.
When to use: any emission or absorption wavelength; for a hydrogen-like ion keep the $Z^2$ factor in front.
[JEE Tip] The shortest wavelength of a series is its limit (largest jump); the longest wavelength is the adjacent-level jump (smallest gap). Swapping these two is the single most common Rydberg slip.

3. Nuclear size, mass defect and binding energy

Nuclear radius: $R=R_0A^{1/3}$ with $R_0=1.2$ fm; the nucleon density, and hence the mass density $\approx2.3\times10^{17}$ kg/m $^3$ , is the same for every nucleus.
Mass-energy: $1$ u $=931.5$ MeV (energy equivalent) $=1.66\times10^{-27}$ kg.
Mass defect: $\Delta m=[Z\,m_p+(A-Z)\,m_n]-m_{\text{nucleus}}$ ; with atomic masses use $Z\,m_H$ so the electrons cancel out.
Binding energy: $BE=\Delta m\,c^2=\Delta m(\text{in u})\times931.5$ MeV; divide by $A$ to get $BE$ per nucleon.
Binding-energy-per-nucleon curve: rises fast, peaks near $8.8$ MeV around $A\approx56$ (iron), then falls slowly. Light nuclei release energy by fusion, heavy nuclei by fission.
When to use: stability comparisons, and any energy-release problem (fusion, fission, decay) once the masses are known.
[JEE Tip] Divide the binding energy by $A$ (all nucleons), never by $Z$ . When the data are atomic masses, use $m_H=1.007825$ u in place of the bare proton and the electron masses book-keep themselves.

4. Radioactivity and Q-value

Decay law: $N=N_0e^{-\lambda t}$ ; the activity $A=A_0e^{-\lambda t}$ falls with the same $\lambda$ .
Activity: $A=\lambda N$ (unit becquerel, $1$ Bq is one decay per second; $1$ Ci $=3.7\times10^{10}$ Bq).
Half-life: $t_{1/2}=\dfrac{0.693}{\lambda}=\tau\ln 2$ .
Mean life: $\tau=\dfrac1\lambda=\dfrac{t_{1/2}}{0.693}=1.44\,t_{1/2}$ , always longer than the half-life.
After $n$ half-lives the surviving fraction is $\left(\dfrac12\right)^n$ with $n=\dfrac{t}{t_{1/2}}$ , which need not be a whole number.
Displacement law: alpha decay lowers $A$ by $4$ and $Z$ by $2$ ; beta-minus leaves $A$ fixed and raises $Z$ by $1$ .
Q-value: $Q=[\,m_{\text{reactants}}-m_{\text{products}}\,]c^2$ ; a positive $Q$ releases energy, and for a parent at rest the products share $Q$ inversely as their masses.
Distance of closest approach (Rutherford): $d=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}=\dfrac{2Z\times1.44}{K(\text{in MeV})}$ fm.
When to use: clean powers of two use the $\left(\dfrac12\right)^n$ form; ratios that are not (say an activity dropping $5000$ to $3000$ ) need the logarithmic $\lambda=\dfrac1t\ln\dfrac{A_0}{A}$ .
[JEE Tip] Keep half-life and mean life apart: $\tau=1.44\,t_{1/2}$ , so the mean life is the longer one. Convert the half-life to seconds before computing an activity in becquerel.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Bohr radius, energy and speed for a hydrogen-like ion. Find the orbital radius, the electron energy and the electron speed in the $n=3$ state of the He $^+$ ion ( $Z=2$ ).

Radius: $r_3=0.529\dfrac{n^2}{Z}=0.529\times\dfrac{9}{2}=2.38$ angstrom.
Energy: $E_3=-13.6\dfrac{Z^2}{n^2}=-13.6\times\dfrac{4}{9}=-6.04$ eV.
Speed: $v_3=2.18\times10^6\times\dfrac{Z}{n}=2.18\times10^6\times\dfrac{2}{3}=1.45\times10^6$ m/s.

Every figure is just the hydrogen value rescaled by the right power of $\dfrac{Z}{n}=\dfrac{2}{3}$ .

Example 2 — Spectral-line wavelength from the Rydberg formula. Find the wavelength of the second Balmer line of hydrogen, the H-beta line from the $n_2=4$ to $n_1=2$ transition. Take $R=1.097\times10^7$ per metre.

Level factor: $\dfrac{1}{2^2}-\dfrac{1}{4^2}=\dfrac14-\dfrac{1}{16}=\dfrac{3}{16}$ .
Wavenumber: $\dfrac1\lambda=R\times\dfrac{3}{16}=1.097\times10^7\times0.1875=2.057\times10^6$ per metre.
Wavelength: $\lambda=4.86\times10^{-7}$ m, i.e. $486$ nm (blue-green, in the visible band).

The H-alpha line ( $3\to2$ ) sits at $656$ nm; this next Balmer line is shorter because its energy gap is larger.

Example 3 — Counting spectral lines. A gas of hydrogen atoms is excited to the $n=6$ level and de-excites by every allowed route. How many distinct lines appear in all, and how many belong to the Lyman series?

Total lines from level $n$ : $\dfrac{n(n-1)}{2}=\dfrac{6\times5}{2}=15$ distinct wavelengths.
Lyman lines end on $n_1=1$ , arriving from $n_2=2,3,4,5,6$ , which is $5$ lines.

So $15$ lines in all, of which $5$ are Lyman (UV); the remaining $10$ split among the Balmer, Paschen and higher series.

Example 4 — Binding energy per nucleon of the alpha particle. Compute the total binding energy and the binding energy per nucleon of $^{4}_{2}\mathrm{He}$ . Use $m_H=1.007825$ u, $m_n=1.008665$ u, $m(^4\mathrm{He})=4.002603$ u and $1$ u $=931.5$ MeV.

Mass defect: $\Delta m=2m_H+2m_n-m(^4\mathrm{He})=4.032980-4.002603=0.030377$ u.
Binding energy: $BE=0.030377\times931.5=28.3$ MeV.
Per nucleon: $\dfrac{BE}{A}=\dfrac{28.3}{4}=7.07$ MeV.

At $7.07$ MeV per nucleon the alpha is exceptionally tightly bound for so light a nucleus, which is why it survives intact and is ejected whole in alpha decay.

Example 5 — Q-value of a nuclear reaction. Find the energy released in $^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{H}\rightarrow2\,^{4}_{2}\mathrm{He}$ . Take $m(^7\mathrm{Li})=7.016004$ u, $m(^1\mathrm{H})=1.007825$ u, $m(^4\mathrm{He})=4.002603$ u and $1$ u $=931.5$ MeV.

Reactant mass: $7.016004+1.007825=8.023829$ u.
Product mass: $2\times4.002603=8.005206$ u.
Mass lost: $\Delta m=0.018623$ u, so $Q=0.018623\times931.5=17.3$ MeV.

Because $Q>0$ the reaction is exoergic; this is the historic Cockcroft-Walton disintegration, the first nucleus split by man-made accelerated protons.

Example 6 — Decay constant, activity and fraction remaining. A source of half-life $30$ days initially holds $6.0\times10^{18}$ nuclei. Find the decay constant, the initial activity, and the fraction left after $120$ days. Take $1$ day $=86400$ s.

Decay constant: $t_{1/2}=30\times86400=2.592\times10^6$ s, so $\lambda=\dfrac{0.693}{2.592\times10^6}=2.67\times10^{-7}$ per second.
Initial activity: $A=\lambda N=(2.67\times10^{-7})(6.0\times10^{18})=1.60\times10^{12}$ Bq.
Fraction left: $120$ days is $n=\dfrac{120}{30}=4$ half-lives, so a fraction $\left(\dfrac12\right)^4=\dfrac{1}{16}=0.0625$ survives.

Note the half-life had to be converted to seconds before the activity came out in becquerel.

Example 7 — Mean life, half-life and the exponential law. A nuclide has a mean life of $20$ days. Find its decay constant, its half-life, and the fraction of nuclei still present after $40$ days. Take $\ln 2=0.693$ .

Decay constant: $\lambda=\dfrac1\tau=\dfrac{1}{20}=0.05$ per day.
Half-life: $t_{1/2}=0.693\,\tau=0.693\times20=13.9$ days, shorter than the mean life.
Survivors: after $40$ days $\lambda t=0.05\times40=2$ , so $\dfrac{N}{N_0}=e^{-\lambda t}=e^{-2}=0.135$ .

About $0.135$ of the sample is left; since $40$ days is not a whole number of half-lives, the exponential form, not $\left(\dfrac12\right)^n$ , is the clean route.