Question bank — Hydrogen energy levels Eₙ = −13.6 - n² eV

True — lekin "zyaada" ka matlab hai kam negative, yaani $0$ ke kareeb. Kyunki $E_n=-13.6/n^2$ hai, bada $n$ fraction ko shrink karta hai, isliye electron well mein upar hota hai (zyaada weakly bound), neeche nahi.

True — $n=1$ deta hai $-13.6$ eV, sabse gehri shelf. Koi bhi allowed state uske neeche nahi hoti; ye ground state hai aur isliye hydrogen stable hai.

False — koi bhi positive total energy matlab hai electron free ho chuka hai (escape kar gaya) aur uske paas leftover kinetic energy hai. Bound states strictly negative hoti hain; formula sirf negative values produce karta hai.

False — kyunki spacing $1/n^2$ follow karti hai, bottom gap ($1\to 2$) bahut bada hai ($10.2$ eV) aur gaps $n$ badhne ke saath zero ki taraf shrink hoti hain. Levels $E=0$ ke paas ek saath crowd ho jaate hain (upar ke well diagram mein dekho).

True — $n$ saare positive integers par chalta hai, aur $-13.6/n^2$ kabhi bhi $0$ tak pahunche bina uss ke kareeb aata jaata hai, isliye infinitely many levels escape line ke thoda neeche ikatte ho jaate hain.

False — wo constant hydrogen ke liye hai ($Z=1$). Neutral helium mein do electrons hain aur alag structure hai; sirf ek one-electron ion jaise He${}^{+}$ hi $E_n=-13.6Z^2/n^2$ ko $Z=2$ ke saath follow karta hai.

True — force balance $m{v}^2=k{e}^2/r$ se (jahan $k=1/4\pi {\epsilon }_0$), kinetic energy $\frac{1}{2}k{e}^2/r$ hai aur potential $-k{e}^2/r$ hai, isliye unka sum $-\frac{1}{2}k{e}^2/r=-\text{KE}$ hai. Isliye bound states automatically negative aate hain.

True — depth (zero se kitni neeche) $13.6/n^2$ hai; $n$ ki jagah $2n$ rakhne par $1/4$ se multiply ho jaata hai. Toh $n=2$, $n=1$ se ek quarter depth par hai.

False — ground state sabse lowest shelf hai, isliye aur neeche girne ki jagah nahi hai. Emission ke liye lower level par girna padta hai, jo ground state ke paas nahi hoti.

Error: divide karne ki jagah multiply kar diya. Rule hai $-13.6/n^2$, toh $E_3=-13.6/9\approx -1.51$ eV. Upar wali shelves shallower hoti hain, deeper nahi.

Error: ek photon do levels ka difference carry karta hai, na ki ek level ki value. Use karo $\Delta E=E_3-E_2=(-1.51)-(-3.40)=1.89$ eV; aur released energy positive hoti hai.

Error: kinetic energy (KE) indeed positive hai, lekin total energy mein negative potential energy (PE $=-k{e}^2/r$) bhi hai, jo jeet jaati hai. Kisi bhi bound electron ka net total negative hota hai.

Error: $13.6$ eV use ground state se free karta hai. $n=2$ se tum sirf $-3.40$ eV se $0$ tak chadh rahe ho, jiske liye $3.40$ eV chahiye. Ionization energy starting shelf par depend karti hai.

Error: wo rukta nahi — wo slower move karta hai ($v\propto 1/n$) lekin finite $n$ ke liye kabhi zero nahi. Aur "ruka hua aur free" $n\to \infty$ ka limit hai, jahan $E\to 0$ hota hai, na ki ek real finite orbit. (Yahan $a_0=0.529$ Å Bohr radius hai.)

Error: sign ka direction se koi lena-dena nahi. Ye encode karta hai ki state free-electron reference $E=0$ ke neeche hai — ye binding ke baare mein statement hai, motion ke baare mein nahi.

Error: closely-spaced high levels tiny photon energies dete hain. Sabse bade photons sabse wide gap se aate hain — $n=1$ par khatam hone wale transitions (Lyman series).

Kyunki "electron infinitely dur, at rest" Coulomb's law mein potential energy ka natural zero hai. Ye bound states ko negative banata hai aur "use free karne ki energy" simply $-E_n$ ke barabar ho jaati hai.

Force balance $m{v}^2=k{e}^2/r$ kinetic energy ko potential se baandh deta hai: KE $=\frac{1}{2}|\text{PE}|$. KE aur PE ko add karne par half (negative) potential milta hai. Ye inverse-square attraction ka ek general result hai (virial theorem).

Kyunki sirf discrete energy shelves exist karti hain, isliye sirf discrete energy differences exist karti hain, aur har difference ek photon energy se map hoti hai $E=hc/\lambda$ ke through. Discrete gaps discrete wavelengths force karte hain — dekho Hydrogen spectral series.

Classically koi bhi orbit allowed hoti hai aur ek accelerating electron radiate karke spiral in ho jaata. Angular momentum ko quantize karna ($\hslash =h/2\pi$ iska natural unit hai) ek unexplained input tha, jo baad mein wave nature se justify hua, jo stable orbits select karta hai — dekho Quantum numbers.

Ye Rydberg energy hai, sirf fundamental constants $m,e,{\epsilon }_0,h$ se bani hoti hai. $n$-dependence puri tarah $1/n^2$ factor mein rehti hai jo uske bahar hota hai — dekho Rydberg constant and formula.

Girne ka matlab hai zyaada negative (lower) level par jaana, toh starting level upar hoti hai; upar-minus-neeche ka difference positive hota hai. Wo positive amount emitted photon ki energy ban jaata hai.

Kyunki sirf quantized orbits allowed hain; $n=1$ ke neeche koi shelf nahi hai. Ground state ka nonzero radius $a_0=0.529$ Å hai aur wo aur energy lose nahi kar sakta, isliye wo stable hai.

$13.6/n^2\to 0$, toh $E_n\to 0^{-}$: electron freedom ki bilkul brink par hai, infinitely weakly bound, infinite radius par. Ye limit hi ionization threshold hai.

Nahi — $n$ shuru hota hai $1$ se. $n=0$ set karne par zero se divide hoga (infinite depth) aur ye $mvr=n\hslash$ bhi violate karega, jisse zero angular momentum milegi. Sabse lowest real shelf $n=1$ hai.

Electron escape kar jaata hai aur surplus kinetic energy rakh leta hai. $E=0$ se upar energies continuous hoti hain, isliye koi bhi excess allowed hai — discrete bound region ke unlike jo neeche hoti hai.

Nahi — ek bound-to-bound jump ke liye aisi energy chahiye jo exactly ek gap se match kare. Ek mismatched photon bina chhua guzar jaata hai (jab tak ki wo puri $13.6$ eV exceed na kare aur ionize na kare).

$E_1=-13.6\times 2^2=-54.4$ eV, hydrogen se chaar guna deeper kyunki doubled nuclear charge single electron ko bahut zyaada khinchti hai ($E\propto Z^2$).

$n=1$ par khatam hona sabse wide gaps cross karta hai, jo sabse energetic (ultraviolet, Lyman) photons deta hai; $n=2$ par khatam hona visible Balmer lines deta hai. Final shelf series aur rough colour band set karta hai.