Exercises — Hydrogen energy levels Eₙ = −13.6 - n² eV

KYA karna hai: $n$ ko $E_n=-13.6/n^2$ mein substitute karo.

• $E_1=-13.6/1^2=\boxed{-13.6\text{eV}}$
• $E_2=-13.6/2^2=-13.6/4=\boxed{-3.40\text{eV}}$

Negative kyun: humne energy ko free-electron mark $E(\infty )=0$ se measure kiya; ek bound electron us mark ke neeche hota hai, isliye uski energy negative hai. $n=1$ sabse gehri shelf hai.

JHOOTH. Jaise $n$ badhta hai, $1/n^2$ ghatkar $0$ ki taraf jaata hai, toh $E_n=-13.6/n^2$ upar $0$ ki taraf chadtha hai (kam negative). Electron upper shelves par zyada loosely bound hota hai.

Quick check: $E_1=-13.6$, $E_2=-3.40$, $E_3=-1.51$ — numbers $0$ ki taraf upar jaate hain. ✓

Ionization = electron ko uski shelf se upar free mark $E=0$ tak kheenchna. $\Delta E=0-E_1=0-(-13.6)=\boxed{13.6\text{eV}}$ Isliye hi number $13.6$ har jagah aata hai — yeh bottom shelf ki gehraai hai.

KYA & KYU: ek photon do shelves ka fark carry karta hai (energy conserved hoti hai). $\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6\cdot \frac{5}{36}=\boxed{1.89\text{eV}}$ $hc=1240\text{eV⋅nm}$ se convert karo (ek division eV ko nm mein badal deta hai): $\lambda =\frac{1240}{1.89}\approx \boxed{656\text{nm}}(\text{red — H-alpha Balmer line})$

$\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\cdot \frac{3}{4}=10.2\text{eV}$ $\lambda =\frac{1240}{10.2}\approx \boxed{122\text{nm}}(\text{ultraviolet})$ UV kyun: yeh poore atom ka sabse bada gap hai (gehre basement mein), isliye photon bahut energetic hai → bahut chhoti wavelength → violet se pare, UV mein.

Electron $E_1=-13.6$ eV par shuru hota hai. Absorb karne ke baad: $E_{\text{final}}=-13.6+12.09=-1.51\text{eV}$ Kaun si shelf $-1.51$ par hai? Solve karo $-13.6/n^2=-1.51\Rightarrow n^2=13.6/1.51=9\Rightarrow \boxed{n=3}$. Yeh kyun kaam karta hai: photon ki energy exactly ek shelf-to-shelf gap se match karni chahiye, warna woh absorb hi nahi hogi. $12.09$ eV precisely $1\to 3$ ka gap hai.

$E_2-E_1=-3.40-(-13.6)=10.2\text{eV}$ $E_3-E_2=-1.51-(-3.40)=1.89\text{eV}$ $E_4-E_3=-0.85-(-1.51)=0.66\text{eV}$ Pattern: gaps tezi se ghatate hain — $10.2\to 1.89\to 0.66$. Shelves $E=0$ ki taraf bheed jaate hain kyunki $1/n^2$ flat ho jaata hai.

Neeche di gayi figure kaise padhein. Vertical axis energy eV mein hai, sahi scale par — jo shelf do guni gehri hai woh do guni neeche draw ki gayi hai. Har solid white horizontal line ek allowed shelf $n=1,\dots ,5$ hai; sabse upar dashed pink line $E=0$ hai, "free electron" mark. Left par, blue numbers har shelf ki energy $E_n$ dete hain (wohi values jo tumne abhi compute kiye, $-13.6,-3.40,-1.51,\dots$). Right par, yellow labels level number $n$ dete hain. Yellow double-arrow bottom gap $E_2-E_1=10.2$ eV measure karta hai — board par sabse bada jump. Pink double-arrow top gap $E_4-E_3=0.66$ eV measure karta hai — barely ek sliver. Toh do arrows literally tumhare algebra ke do numbers hain, heights ke roop mein draw kiye gaye: yellow arrow pink se lagbhag $15\times$ lamba hai, jo exactly $10.2/0.66$ hai. Kya observe karein: $n$ mein equal steps lene se energy mein wildly unequal steps milte hain — yahi hydrogen spectrum ki poori kahani hai.

Yeh tool kyun: radii $r_n=a_0{n}^2$ ke hisaab se scale hoti hain (parent mein force balance + quantization se derive kiya gaya hai).

• $r_1=a_0\cdot 1^2=0.0529\text{nm}$
• $r_3=a_0\cdot 3^2=9a_0=0.476\text{nm}$
• Factor: $r_3/r_1=9$. $\boxed{r_3=9r_1=0.476\text{nm}}$. Upar ki shelves bahut zyada door hoti hain — isliye woh weakly bound hoti hain.

Predict: $n=1$ mein koi bhi jump bottom ka bada gap cross karta hai, toh $3\to 1$ jeetna chahiye. Verify: $\Delta E_{4\to 2}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6\cdot \frac{3}{16}=2.55\text{eV}$ $\Delta E_{3\to 1}=13.6\left(1-\frac{1}{9}\right)=13.6\cdot \frac{8}{9}=12.09\text{eV}$ $12.09>2.55$ ✓ — $\boxed{3\to 1}$ kaafi zyada energetic hai (ek Lyman/UV line).

$n=\infty$ kya hai? Electron free shuru hota hai ($E_{\infty }=0$) aur $n=2$ par girta hai. $\Delta E=13.6\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)=13.6\cdot \frac{1}{4}=3.40\text{eV}$ $\lambda =\frac{1240}{3.40}\approx \boxed{365\text{nm}}$ "Limit" kyun: jaise $n_i\to \infty$ hota hai, gaps badhna band ho jaate hain, toh photon energy $3.40$ eV par cap ho jaati hai — yeh maximum energy (isliye minimum $\lambda$) hai jo koi bhi Balmer line reach kar sakti hai.

$Z^2$ kyun: proton ka pull $Z$ guna zyada strong hota hai, aur energy charge squared carry karta hai: $E_n=-13.6Z^2/n^2$. $E_1({\text{He}}^{+})=-13.6\cdot \frac{2^2}{1^2}=-13.6\cdot 4=\boxed{-54.4\text{eV}}$ Ionization energy $=|E_1|=\boxed{54.4\text{eV}}$ — hydrogen ka chaar guna, kyunki nucleus apne ek electron ko bahut zyada tightly pakad ke rakhta hai.

$\Delta E=13.6Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=13.6\cdot 4\cdot \frac{3}{4}=40.8\text{eV}$ $\lambda =\frac{1240}{40.8}\approx \boxed{30.4\text{nm}}$ H ke $122$ nm se compare karo, yeh exactly $4\times$ chhota hai ($122/4=30.5$). KYU: har energy $Z^2=4$ se scale hoti hai, toh har wavelength $4$ se shrink hoti hai. Same pattern, $Z^2$ se "zoom in" hua.

Step 1 — photon ki energy: $E=\frac{1240}{434}=2.857\text{eV}$ Step 2 — series guess karo. Visible light (Balmer) matlab $n_f=2$. Test karo: $2.857=13.6\left(\frac{1}{4}-\frac{1}{n_i^2}\right)\Rightarrow \frac{1}{n_i^2}=\frac{1}{4}-\frac{2.857}{13.6}=0.25-0.210=0.040$ $n_i^2=\frac{1}{0.040}=25\Rightarrow n_i=5$ Answer: $\boxed{n=5\to n=2}$ (H-gamma Balmer line). ✓ $n_f=2$ se kyun shuru karein: sirf Balmer lines visible band mein aati hain; pehle series fix karne se do-unknown problem ek-line solve ban jaati hai.

Symbols jo hum use karenge (yahan define kiye taaki kuch unexplained na ho):

• $e$ = electron ki charge ki size (proton ki bhi), $e=1.602\times 10^{-19}\text{C}$ (coulombs). Yeh set karta hai ki proton aur electron kitna strongly attract karte hain.
• ${\epsilon }_0$ = vacuum permittivity, ek fixed constant of nature, ${\epsilon }_0=8.854\times 10^{-12}$ (SI units mein). Yeh Coulomb's law mein aata hai aur electric force ki strength-scale fix karta hai. Hum ise ek symbol mein bundle karte hain $k=\frac{1}{4\pi {\epsilon }_0}=8.99\times 10^9$ taaki algebra clean rahe.
• $r$ = orbit radius, $v$ = electron speed.

Step 0a — do forces alag-alag likho. Proton electron ko Coulomb's law se inward force se attract karta hai $F_{\text{Coulomb}}=k\frac{e^2}{r^2}\left(k=\frac{1}{4\pi {\epsilon }_0}\right).$ Circle mein move karte rehne ke liye electron ko ek inward Centripetal force chahiye $F_{\text{centripetal}}=\frac{m{v}^2}{r}.$ Step 0b — dono ko equal karo (yahi poora Bohr idea hai). Coulomb pull hi centripetal force hai — koi aur inward push nahi — toh $k\frac{e^2}{r^2}=\frac{m{v}^2}{r}.$ Step 0c — ek $r$ cancel karo aur dekho KYU. Left side mein denominator mein $r^2$ hai, right side mein sirf $r$. Dono sides ko $r$ se multiply karo: left side ek power kho deta hai ($r^2\to r$), right side ka denominator completely khatam ho jaata hai: $k\frac{e^2}{r}=m{v}^2.$ Yeh ek cancellation hi agli line ko clean banata hai. Step 1 — kinetic energy padho. Kinetic energy $\text{KE}=\frac{1}{2}m{v}^2$ hai, aur humne abhi nikala $m{v}^2=k{e}^2/r$, toh $\text{KE}=\frac{1}{2}k\frac{e^2}{r}.$ Step 2 — potential energy. Free electron $E(\infty )=0$ ko reference lete hue, electric potential energy hai $\text{PE}=-k\frac{e^2}{r}=-2\text{KE}.$ Step 3 — dono add karo. $E=\text{KE}+\text{PE}=\text{KE}-2\text{KE}=-\text{KE}.$ Toh total energy exactly minus kinetic energy hai — yahi on-page justification hai. Step 4 — $n=1$ par numbers. Kyunki $E_1=-13.6$ eV aur $E=-\text{KE}$, hume milta hai KE $=13.6\text{eV}$. Convert karo aur $\text{KE}=\frac{1}{2}m{v}^2$ invert karo: $\text{KE}=13.6\times 1.602\times 10^{-19}=2.179\times 10^{-18}\text{J}$ $v=\sqrt{\frac{2\text{KE}}{m}}=\sqrt{\frac{2(2.179\times 10^{-18})}{9.11\times 10^{-31}}}=\sqrt{4.78\times 10^{12}}\approx \boxed{2.19\times 10^6\text{m/s}}$ Yeh light speed ka lagbhag $0.7\%$ hai — fast, lekin non-relativistic, toh Bohr picture theek hai.

KYU: kyunki $E_{\text{photon}}=13.6(1/n_f^2-1/n_i^2)$ aur $1/\lambda =E_{\text{photon}}/hc$: $\frac{1}{\lambda }=\frac{13.6}{hc}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ Toh single constant $R_H=13.6/hc$ har line control karta hai — dekho Rydberg constant and formula. $R_H=\frac{13.6\text{eV}}{1240\text{eV⋅nm}}=0.01097{\text{nm}}^{-1}=1.097\times 10^7{\text{m}}^{-1}$ Measured Rydberg constant se match karta hai. ✓ Poora line spectrum ek number hai jo do square-reciprocals ke difference se multiply hota hai.