Class 12 Physics · Chapter 12
Short answer:
Ye chapter 2026-27 ke rationalised syllabus me moderate depth ka hai — Rutherford ka alpha-scattering experiment conceptually poora hai, lekin Bohr model ab sirf FINAL orbit-radius aur energy expressions use karne ke liye hai, poora step-by-step quantization derivation nahi karna. Spectral series (Lyman, Balmer, Paschen) ki detailed wavelength-formula depth bhi hata di gayi hai — bas itna samajhna hai ki electron transitions se spectral lines banti hain. Neeche 14 exercise-style questions hain: Rutherford scattering conceptual, Bohr radius/energy calculation, transition energy (release/absorb), aur ionization energy — sabme moderate numericals hain, heavy derivation nahi.
Rutherford ka alpha-scattering experiment prove karta hai ki atom ka zyada tar mass aur poora positive charge ek chhote nucleus me concentrated hai — kyunki kuch alpha particles bahut bade angle se bounce back ho gaye, jo tabhi possible hai jab andar kuch bahut dense ho. Isi observation ne Rutherford ke nuclear model ko janam diya, lekin us model ki ek badi kami thi — classical physics ke hisaab se orbit me ghoomta electron continuously radiation kho kar nucleus me gir jaana chahiye tha. Bohr ne ye problem fix kiya by proposing ki electron sirf kuch fixed (quantized) orbits me hi stable rehta hai, aur is chapter me humein Bohr model ki poori derivation nahi, sirf uske FINAL orbit-radius aur energy formulas use karne aate hain.
Chapter 12 Summary — 5 Minute Revision
1. Rutherford ka alpha-particle scattering experiment
Rutherford aur unke students ne ek thin gold foil (sona ki patli sheet) pe fast-moving alpha particles (helium nucleus, positively charged) ki beam fire ki aur unka scattering pattern study kiya.
- Observation 1: Zyada tar alpha particles bina kisi deflection ke seedha foil se guzar gaye — iska matlab atom ka zyada tar volume khaali (empty space) hai.
- Observation 2: Kuch alpha particles small angles se deflect hue — ye batata hai ki atom ke andar kahin electric charge concentrated hai jo unhe repel karta hai.
- Observation 3 (sabse surprising): Bahut kam (~1 in 8000) alpha particles 90° se bhi zyada deflect hue, kuch to almost 180° (wapas bounce) kar gaye. Ye sirf tabhi possible hai jab atom ke andar ek bahut chhota, bahut dense, positively charged region ho jo poore atom ka mass aur positive charge apne andar concentrate kiye ho — isi region ko Rutherford ne nucleus naam diya.
Conclusion — Rutherford's nuclear model of atom:
- Atom ka zyada tar mass aur poora positive charge ek bahut chhote, dense nucleus me concentrated hota hai (nucleus ka size atom ke size se ~10⁴-10⁵ guna chhota hai).
- Electrons nucleus ke charge se bahut dur, khaali space me orbit karte hain — isliye atom mostly empty space hai.
- Nucleus ke around electrons circular orbits me revolve karte hain, jaise sun ke around planets — isliye ise planetary model bhi kehte hain.
Rutherford model ki limitation: Classical electromagnetic theory ke hisaab se, ek accelerating charge (jaise orbit me ghoomta electron) continuously energy radiate karta hai. Agar aisa hota, to electron dheere-dheere energy khokar spiral path me nucleus ke andar gir jaata, aur atom stable hi nahi reh pata — jo observation ke bilkul opposite hai (atoms toh stable hote hain). Isi contradiction ko solve karne ke liye Bohr model aaya.
2. Bohr model of the hydrogen atom — final results
Bohr ne propose kiya ki electron sirf kuch specific (quantized) circular orbits me hi stable reh sakta hai, jinme woh energy radiate nahi karta. In orbits ko principal quantum number n (n = 1, 2, 3, ...) se label kiya jaata hai. Is chapter me humein poori derivation nahi karni — sirf FINAL results use karne aane chahiye:
rₙ ∝ n² (r₁ = 0.529 Å, hydrogen ke ground state ka orbit radius)
Yahan r₁ = 0.529 Å (1 Å = 10⁻¹⁰ m) hydrogen atom ke first orbit (ground state, n=1) ka radius hai. Isliye n-th orbit ka radius:
rₙ = n² × 0.529 Å
Matlab orbit radius n² ke proportion me badhta hai — n=2 pe radius 4 guna, n=3 pe 9 guna ho jaata hai.
Similarly, n-th orbit me electron ki total energy:
Eₙ = −13.6 / n² eV
Negative sign ka matlab electron nucleus se bound hai (energy zero tab hoti jab electron atom se poori tarah free/infinite door ho jaaye). Ground state (n=1) me energy sabse kam (most negative) hoti hai — E₁ = −13.6 eV — aur jaise-jaise n badhta hai, energy zero ki taraf badhti hai (less negative, matlab electron kam tightly bound hota hai).
3. Energy level diagram aur transitions
- Electron kisi bhi allowed orbit (energy level) me tab tak reh sakta hai jab tak koi external energy na mile — ise stationary state kehte hain.
- Jab electron higher energy level se lower level me jump karta hai, to energy difference ke barabar ek photon release hota hai.
- Jab electron lower level se higher level me jaata hai, to usko utni hi energy absorb karni padti hai (kisi external source se, jaise photon ya collision).
- Released/absorbed photon ki energy = do levels ki energy ka difference: ΔE = |E_final − E_initial|.
Sabse kam energy wali state (n=1) ko ground state kehte hain, aur usse upar ki koi bhi state (n=2,3,...) ko excited state kehte hain.
4. Atomic spectra — basic idea
Jab hydrogen gas ko excite kiya jaata hai (heat ya electric discharge se), electrons higher levels me chale jaate hain aur phir wapas lower levels me girte hue photons emit karte hain. Ye photons specific, fixed wavelengths ke hote hain (kyunki energy levels khud quantized/fixed hain) — isliye continuous spectrum ki jagah discrete spectral lines dikhti hain jab isse prism se guzara jaata hai. Har spectral line ek specific transition (specific pair of n values) ko represent karti hai. Is syllabus me detailed series names (Lyman, Balmer, Paschen) ya unki quantitative wavelength formulas nahi padhni — bas itna samajhna hai ki discrete energy levels hi discrete spectral lines ki wajah hain.
5. Ionization energy
Ground state (n=1) me bound electron ko poori tarah atom se nikalkar free (n=∞, jahan energy zero hai) karne ke liye jo minimum energy chahiye, use ionization energy kehte hain. Hydrogen atom ke liye:
Ionization energy = E∞ − E₁ = 0 − (−13.6 eV) = 13.6 eV
Poore Class 12 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q14)
Q1. Rutherford ke alpha-scattering experiment me zyada tar alpha particles bina deflect hue seedhe foil se kyun guzar gaye?
Kyunki atom ka zyada tar volume khaali space hota hai. Positive charge aur mass sirf ek bahut chhote nucleus me concentrated hote hain, isliye zyada tar alpha particles ko rasta me koi significant obstruction nahi milti aur woh seedhe guzar jaate hain.
Q2. Kuch alpha particles 90° se bhi zyada angle pe deflect (kabhi almost 180°) kyun hue?
Jab koi alpha particle bahut close se nucleus ke paas se guzarta hai, to nucleus ka poora positive charge (jo ek bahut chhoti jagah me concentrated hai) usse strongly repel karta hai. Ye strong repulsive force alpha particle ko bade angle se, kabhi-kabhi almost seedha wapas, deflect kar deti hai. Ye tabhi possible hai jab atom ke andar koi bahut dense, chhota, positively charged region ho.
Q3. Rutherford ke observations se atom ki structure ke baare me kya conclusions nikale gaye?
(i) Atom ka zyada tar volume khaali hota hai. (ii) Atom ka poora positive charge aur zyada tar mass ek bahut chhote, dense nucleus me concentrated hota hai. (iii) Electrons is nucleus ke around, khaali space me, orbits me revolve karte hain (planetary model).
Q4. Classical electromagnetic theory ke hisaab se Rutherford ka model kyun unstable tha?
Classical theory ke hisaab se, koi bhi accelerating charge (jaise orbit me circular motion karta electron) continuously electromagnetic energy radiate karta hai. Agar electron aisa karta rehta, to woh dheere-dheere apni energy kho kar spiral path follow karte hue nucleus me gir jaata — jisse atom collapse ho jaata. Lekin real atoms stable hote hain, isliye ye contradiction Bohr model se resolve kiya gaya.
Q5. Hydrogen atom ke second orbit (n=2) ka radius calculate karo.
Formula: rₙ = n² × 0.529 Å
r₂ = (2)² × 0.529 Å = 4 × 0.529 Å = 2.116 Å
Q6. Hydrogen atom ke third orbit (n=3) ka radius calculate karo, meters me.
rₙ = n² × 0.529 Å, aur 1 Å = 10⁻¹⁰ m
r₃ = 9 × 0.529 Å = 4.761 Å = 4.761 × 10⁻¹⁰ m
Q7. Hydrogen atom ki ground state (n=1) energy kya hai, aur uska sign negative kyun hai?
E₁ = −13.6/1² = −13.6 eV. Sign negative isliye hai kyunki electron nucleus se bound (attracted) hai — energy zero tab hoti jab electron atom se poori tarah free (infinitely door) ho jaaye. Negative energy is bound state ko represent karti hai.
Q8. Hydrogen atom ki n=4 state ki energy calculate karo.
Formula: Eₙ = −13.6/n² eV
E₄ = −13.6/4² = −13.6/16 = −0.85 eV
Q9. Electron n=4 se n=2 me transition karta hai. Is transition me energy release hogi ya absorb, aur kitni?
Electron higher level (n=4) se lower level (n=2) me ja raha hai, isliye energy release hogi.
E₄ = −13.6/16 = −0.85 eV, E₂ = −13.6/4 = −3.4 eV
Released energy = E₄ − E₂ = (−0.85) − (−3.4) = 2.55 eV
Q10. Electron n=1 se n=3 me jump karne ke liye kitni energy absorb karni hogi?
Electron lower level (n=1) se higher level (n=3) me ja raha hai, isliye energy absorb hogi.
E₁ = −13.6 eV, E₃ = −13.6/9 = −1.51 eV
Absorbed energy = E₃ − E₁ = (−1.51) − (−13.6) = 12.09 eV
Q11. Hydrogen atom ki ionization energy kya hai, aur ye kaise define hoti hai?
Ionization energy = ground state (n=1) me bound electron ko atom se poori tarah nikalkar free (n=∞, energy zero) karne ke liye chahiye minimum energy.
Ionization energy = E∞ − E₁ = 0 − (−13.6 eV) = 13.6 eV
Q12. Ek hydrogen atom ground state me hai. Usse ionize karne ke liye 10 eV energy di jaati hai — kya ye kaafi hai? Reasoning do.
Nahi, kaafi nahi hai. Hydrogen ki ionization energy 13.6 eV hai, jabki di gayi energy sirf 10 eV hai — jo required energy se kam hai. Electron ionize nahi hoga; ho sakta hai ye energy use kisi higher discrete level tak excite kar de agar wo exactly kisi transition energy ke barabar ho, warna atom is energy ko absorb hi nahi karega (kyunki intermediate energies allowed nahi hain).
Q13. Ground state aur excited state me kya farak hai?
Sabse kam energy wali state (n=1, jahan electron nucleus ke sabse pass, sabse tightly bound hota hai) ground state kehlati hai. Isse upar ki koi bhi allowed state (n=2, 3, ...) excited state kehlati hai, jahan electron ne extra energy absorb karke khud ko higher orbit me pahuncha diya hai.
Q14. Discrete (line) atomic spectra kyun banti hain, continuous spectrum kyun nahi?
Electron sirf kuch fixed, quantized energy levels (n=1,2,3,...) me hi reh sakta hai — beech ki koi bhi energy allowed nahi hai. Jab electron ek level se dusre me jump karta hai, to bilkul specific, fixed energy difference ke barabar photon emit/absorb hota hai — matlab specific wavelength ki light. Kyunki energy levels khud discrete hain, unke beech transitions se banne wali spectral lines bhi discrete (fixed wavelengths pe) hoti hain, continuous nahi.
Important Equations — Ek Nazar Me
| Concept | Expression |
|---|---|
| Bohr radius (ground state, hydrogen) | r₁ = 0.529 Å (= 0.529 × 10⁻¹⁰ m) |
| n-th orbit radius | rₙ = n² × 0.529 Å |
| n-th orbit energy (hydrogen) | Eₙ = −13.6 / n² eV |
| Ground state energy | E₁ = −13.6 eV |
| Energy released (transition high → low, n_i → n_f, n_i > n_f) | ΔE = E_i − E_f (positive, photon emitted) |
| Energy absorbed (transition low → high, n_i → n_f, n_f > n_i) | ΔE = E_f − E_i (positive, photon/energy absorbed) |
| Ionization energy of hydrogen | E∞ − E₁ = 0 − (−13.6 eV) = 13.6 eV |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Energy calculation me negative sign bhool jaana. Eₙ = −13.6/n² eV me negative sign zaroori hai — ye batata hai electron bound hai. Isse chhod dena ya galat jagah lagana transition-energy calculations me sign error kar deta hai.
- Radius aur energy dono ko n ke saath same tarah scale samajhna. Radius n² ke proportion me badhta hai (rₙ = n² × 0.529 Å), lekin energy 1/n² ke proportion me badhti hai (Eₙ = −13.6/n² eV) — dono formulas alag hain, inhe swap mat karo.
- Confuse karna ki konsi transition energy release karti hai aur konsi absorb. Higher n se lower n me girna → energy RELEASE (emission). Lower n se higher n me jaana → energy ABSORB. Ye direction galat samajhna sabse common mistake hai.
- Ionization energy ki value galat likhna. Hydrogen ki ionization energy exactly 13.6 eV hoti hai (n=1 se n=∞ tak) — ise 13.6 eV ke alawa koi approximate ya round value likhna galat hai.
- Hydrogen ke Bohr formulas ko bina Z adjust kiye doosre atoms pe apply karna. Ye formulas (rₙ = n² × 0.529 Å aur Eₙ = −13.6/n² eV) sirf hydrogen (Z=1) ke liye hain. Kisi aur atom/ion (jaise He⁺) pe apply karne ke liye atomic number Z ka factor include karna padta hai — bina adjust kiye direct use karna galat hai.
- Poori Bohr derivation likhne ki koshish karna jab sirf final answer chahiye. Current syllabus me detailed quantization argument (angular momentum quantization se radius/energy derive karna) nahi poocha jaata — sirf final rₙ aur Eₙ formulas use karke numerical solve karna hota hai. Poori derivation likhna time waste karta hai aur zaroori bhi nahi hai.
Board-Style Important Questions
- 1 mark: Rutherford ke alpha-scattering experiment se atom ke baare me sabse important conclusion kya nikla?
- 1 mark: Hydrogen atom ki ground state energy kitni hoti hai?
- 2 marks: Kuch alpha particles bade angle se deflect kyun hue the — is observation ka kya matlab nikala gaya?
- 2 marks: Hydrogen atom ke n=3 orbit ka radius aur energy dono calculate karo.
- 3 marks: Electron ek transition me energy release karta hai ya absorb karta hai — ye kaise decide karte hain? Ek numerical example ke saath samjhao.
- 3 marks: Rutherford model classical physics ke hisaab se unstable kyun tha, aur Bohr model ne is problem ko kaise address kiya (qualitatively)?
Aksar Poochhe Jaane Wale Sawaal
Kya humein Bohr model ki poori derivation yaad karni hai?
Nahi. Current rationalised syllabus me sirf FINAL orbit-radius (rₙ = n² × 0.529 Å) aur energy (Eₙ = −13.6/n² eV) expressions use karke numericals solve karne aane chahiye. Angular momentum quantization se poora derivation karna is chapter ka scope nahi hai.
Kya spectral series (Lyman, Balmer, Paschen) ke naam aur formulas yaad karne hain?
Detailed series naming aur unki quantitative wavelength-formula depth is syllabus se hata di gayi hai. Bas itna samajhna kaafi hai ki discrete energy levels ki wajah se discrete spectral lines banti hain — series-specific formulas yaad karna zaroori nahi.
Energy transition me kaise pata karein ki energy release hui ya absorb?
Agar electron higher n se lower n me ja raha hai (jump down), to energy release (emit) hoti hai. Agar lower n se higher n me ja raha hai (jump up), to energy absorb karni padti hai. Direction hamesha check karo before calculating.
Hydrogen ki ionization energy 13.6 eV hi kyun hai, koi aur value kyun nahi?
Ionization energy = E∞ − E₁ = 0 − (−13.6 eV) = 13.6 eV, jahan −13.6 eV hydrogen ke ground state (n=1) ki fixed energy hai. Ye value experimentally aur Bohr model dono se consistent hai, isliye exactly 13.6 eV hi correct hai.
Kya Bohr formulas hydrogen ke alawa doosre atoms pe bhi directly use kar sakte hain?
Nahi, directly nahi. rₙ = n² × 0.529 Å aur Eₙ = −13.6/n² eV sirf hydrogen (Z=1, single electron, single proton) ke liye hain. Kisi doosre single-electron ion (jaise He⁺) ke liye atomic number Z ka factor formula me adjust karna padta hai.
Is chapter ka weightage kitna hai?
Specific mark weightage yahan claim nahi kiya ja sakta kyunki ye exam-to-exam vary karta hai. Concept clear rakho — Rutherford scattering, Bohr model ke final formulas, energy-level transitions, aur ionization energy — yehi is chapter ka core scope hai.
Class 12 Physics — Saare Chapters
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