Exercises — EM spectrum — all bands and applications

Energy frequency ke saath badhti hai, aur frequency is list ko walk karte waqt badhti hai. Toh, low → high energy: $\text{radio}<\text{microwave}<\text{infrared}<\text{visible}<\text{ultraviolet}<\text{X-ray}<\text{gamma}.$ Yeh order kyun? Har step daayein taraf wavelength ko shorten karta hai; ek choti wave har second mein zyada crests deti hai, isliye $\nu$ badhta hai, aur $E=h\nu$ uske saath badhta hai. Ek chain $\lambda \downarrow \Rightarrow \nu \uparrow \Rightarrow E\uparrow$ ke alaawa kuch yaad nahi karna.

• $3\text{cm}=0.03\text{m}$ range $1\text{mm}$–$0.1\text{m}$ mein hai → microwave (radar band).
• $500\text{nm}$ $400$ aur $700\text{nm}$ ke beech hai → visible (blue-green).
• $0.05\text{nm}<0.01\text{nm}$? Nahi — $0.05>0.01$, toh yeh $0.01$–$10\text{nm}$ mein hai → X-ray (atomic-spacing scale ke paas jo crystals ko diffract karti hai).

Kaun sa tool aur kyun? Hume frequency di gayi hai aur length chahiye, aur dono Wave speed c = νλ se locked hain. $\lambda$ ke liye rearrange karo: $\lambda =\frac{c}{\nu }=\frac{3\times 10^8\text{m/s}}{5\times 10^9\text{Hz}}=6\times 10^{-2}\text{m}=6\text{cm}.$ Sanity check: $6\text{cm}$ microwave band mein land karta hai — exactly wahi jahan Wi-Fi rehta hai.

Kaun sa tool aur kyun? Hume wavelength se per-photon energy chahiye, toh Photon energy E = hν ko uske wavelength form $E=hc/\lambda$ mein use karo (kyunki $\nu =c/\lambda$). $E=\frac{hc}{\lambda }=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{650\times 10^{-9}\text{m}}=3.06\times 10^{-19}\text{J}.$ eV mein convert karo "one-eV in joules" exchange rate se divide karke: $E=\frac{3.06\times 10^{-19}\text{J}}{1.6\times 10^{-19}\text{J/eV}}\approx 1.91\text{eV}.$ Sanity check: kuch eV Atomic spectra ka scale hai — yahi wajah hai ki atoms visible light emit karte hain.

Total energy $=$ (photons) $\times$ (energy per photon). Pehle per-photon energy: $E_1=\frac{hc}{\lambda }=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{550\times 10^{-9}\text{m}}=3.616\times 10^{-19}\text{J}.$ Phir pulse energy ko ek photon ki worth se divide karo: $N=\frac{2.0\times 10^{-3}\text{J}}{3.616\times 10^{-19}\text{J}}\approx 5.5\times 10^{15}\text{photons}.$ Divide kyun? Har photon ek indivisible lump hai; count simply yeh hai ki total mein kitne lumps fit hote hain.

Kyunki $E=hc/\lambda$, constants $hc$ ratio mein cancel ho jaate hain — yahi wajah hai ki ratios fast route hain: $\frac{E_A}{E_B}=\frac{hc/{\lambda }_A}{hc/{\lambda }_B}=\frac{{\lambda }_B}{{\lambda }_A}.$ Dono wavelengths ko metres mein rakho: ${\lambda }_A=0.1\times 10^{-9}\text{m}=1\times 10^{-10}\text{m}$, aur ${\lambda }_B=10\times 10^{-6}\text{m}=1\times 10^{-5}\text{m}$. $\frac{E_A}{E_B}=\frac{1\times 10^{-5}\text{m}}{1\times 10^{-10}\text{m}}=1\times 10^5.$ Interpretation: X-ray photon $100,000\times$ zyada energetic hai — itna ki atoms ko ionise kar sake aur soft tissue se pass ho sake, jabki IR photon sirf molecules ko vibrate karta hai (warmth). Same tarah ki wave, wildly different consequence.

"Sabse lamba wavelength jo phir bhi kaam kare" = "exactly threshold energy", kyunki lamba $\lambda$ matlab kam energy jo fail ho jaayegi. Toh $E=5\text{eV}$ set karo aur $E=hc/\lambda$ ko $\lambda$ ke liye solve karo. Pehle joules mein convert karo: $E=5\text{eV}\times 1.6\times 10^{-19}\text{J/eV}=8.0\times 10^{-19}\text{J}$. $\lambda =\frac{hc}{E}=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{8.0\times 10^{-19}\text{J}}\approx 2.49\times 10^{-7}\text{m}=249\text{nm}.$ Kaun sa band? $249\text{nm}$ $400\text{nm}$ se below hai → ultraviolet. Yahi precisely wajah hai ki UV sterilise karta hai (DNA bonds todata hai) aur visible light nahi karta.

Neeche ki figure $E=hc/\lambda$ ko log axes pe plot karti hai. Blue curve ko down-and-right follow karo: jaise wavelength badhti hai, energy girti hai. Yellow dashed line $5\text{eV}$ bond threshold mark karti hai; yeh curve se (green dot pe) exactly us red dashed wavelength pe milti hai, $\approx 249\text{nm}$ — shaded UV aur visible bands ke beech ki border. Us red line ke daayein kuch bhi bond todne ke liye bahut weak hai.

Kaun sa tool aur kyun? Bragg's law successive planes se bounce karne wali waves ke beech path-length difference ko wavelength se relate karta hai; constructive interference ke liye us extra path ko wavelengths ka whole number hona chahiye. $\sin \theta$ ke liye solve karo: $\sin \theta =\frac{n\lambda }{2d}=\frac{(1)(0.154\text{nm})}{2(0.282\text{nm})}=\frac{0.154}{0.564}=\mathrm{0.2730.}$ Phir $\theta =\arcsin (0.2730)\approx 15.85^{\circ }$. Arcsin kyun? Hume sine (ek ratio) pata hai aur woh angle chahiye jo ise produce karta hai — arcsin exactly "kaun sa angle is sine ko rakhta hai?" undo-operation hai.

(a) $\lambda =c/\nu =\frac{3\times 10^8\text{m/s}}{90\times 10^6\text{Hz}}=3.33\text{m}$. Uska quarter hai $\frac{3.33\text{m}}{4}=0.833\text{m}\approx 83\text{cm}$. (Yahi wajah hai ki real FM whip antennas order-of-a-metre long hote hain.) (b) ${\lambda }_{\text{oven}}=\frac{3\times 10^8\text{m/s}}{2.45\times 10^9\text{Hz}}=0.1224\text{m}\approx 12.2\text{cm}$. (c) Energy ratio $=\frac{{\lambda }_{\text{oven}}}{{\lambda }_{\text{vis}}}=\frac{0.1224\text{m}}{500\times 10^{-9}\text{m}}=\frac{0.1224}{5\times 10^{-7}}\approx 2.45\times 10^5$. Toh ek single green photon lagbhag $245,000\times$ zyada energetic hai ek microwave-oven photon se — phir bhi oven khaana pakhaata hai aur green light nahi, kyunki khaana pakhaana total power delivered aur dielectric relaxation ke baare mein hai, per-photon punch ke baare mein nahi.

Oven photon: $E=h\nu =(6.63\times 10^{-34}\text{J⋅s})(2.45\times 10^9\text{Hz})=1.62\times 10^{-24}\text{J}$. eV mein: $E=\frac{1.62\times 10^{-24}\text{J}}{1.6\times 10^{-19}\text{J/eV}}=1.01\times 10^{-5}\text{eV}$. Transition se compare karo: $\frac{2\times 10^{-3}\text{eV}}{1.01\times 10^{-5}\text{eV}}\approx 198$. Transition ko roughly $200\times$ zyada energy chahiye jitni ek oven photon supply karta hai. Conclusion: oven ek molecule ko us gap ke paar ek photon se nahi utha sakta — yeh confirm karta hai ki mechanism bulk dielectric relaxation hai (field polar molecules ko drag karta hai aur unka lag heat dump karta hai), discrete resonance nahi. Yeh exactly parent note ki steel-manned mistake hai, quantitative bana di gayi.

(a) Refractive index $n=c/v$ se define hota hai, toh $v=c/n=\frac{3\times 10^8\text{m/s}}{1.53}=1.96\times 10^8\text{m/s}$. Vacuum se slower — jaisa $n>1$ ke liye hona chahiye. (b) Frequency source se set hoti hai aur medium mein enter karte waqt change nahi hoti (wave crests boundary pe utni hi rate se arrive karte hain jitni rate se leave karte hain). $v=\nu {\lambda }_{\text{glass}}$ aur $\nu =c/{\lambda }_0$ ke saath: ${\lambda }_{\text{glass}}=\frac{v}{\nu }=\frac{{\lambda }_0}{n}=\frac{400\text{nm}}{1.53}\approx 261\text{nm}.$ (c) Nahi — frequency unchanged hai (yahi wajah hai ki colour preserve hota hai). (d) Nahi — $E=h\nu$, aur $\nu$ unchanged hai, toh photon energy glass aur vacuum mein identical hai. Sirf speed aur wavelength shrink hote hain. Yeh parent-note mistake ka resolution hai "different bands different speeds pe travel karte hain": speed changes media mein hoti hain, aur tab bhi energy fixed rehti hai.

Kaun sa tool aur kyun? Hume ek temperature di gayi hai aur poochha gaya hai ki emission sabse strong kahan hai — yeh exactly woh question hai jo Wien's law answer karta hai. Yeh kehta hai ki peak wavelength temperature ke inversely proportional hai, ${\lambda }_{\text{peak}}=b/T$: hotter body ⇒ chota peak wavelength. Toh hum simply har $T$ ko us formula mein feed karte hain. (a) ${\lambda }_{\text{peak}}=\frac{2.90\times 10^{-3}\text{m⋅K}}{5800\text{K}}=5.0\times 10^{-7}\text{m}=500\text{nm}$ → visible (green-ish). (b) ${\lambda }_{\text{peak}}=\frac{2.90\times 10^{-3}\text{m⋅K}}{310\text{K}}=9.35\times 10^{-6}\text{m}\approx 9.4\mu \text{m}$ → infrared. (c) Sun visible mein peak karta hai, toh evolution ne hamaari aankhein us band ke liye tune kiya jahan sabse zyada photons arrive karte hain — yahi wajah hai ki "visible" physically kuch special nahi hai, sirf Sun ka peak hai. Room temperature ke paas warm bodies IR mein peak karti hain, toh ek camera jo IR "dekhta" hai woh glow detect karta hai jo har warm object emit karta hai, andheron mein bhi. Ek law, do temperatures, do bands — pure spectrum logic miniature mein.

Neeche ki figure dono blackbody curves ko shared log-wavelength axis pe draw karti hai. Yellow curve (Sun) shaded green visible band ke andar apni dashed line pe peak karti hai; red curve (human) bahut daayein $\approx 9.4\mu \text{m}$ pe peak karti hai, IR mein deep. Directly read off karo: cooler body ⇒ uski poori curve — aur uska peak — longer wavelengths ki taraf rightward slide karti hai.

90 MHz pe antenna ki quarter-wave length? ::: lagbhag 0.83 m Wi-Fi 5 GHz wavelength? ::: 6 cm 650 nm photon energy eV mein? ::: lagbhag 1.91 eV Sabse lamba wavelength jo 5 eV bond tod sake? ::: lagbhag 249 nm (UV) 0.154 nm X-rays ka 0.282 nm planes se Bragg angle? ::: lagbhag 15.85° Kya glass mein enter karte waqt frequency change hoti hai? ::: Nahi — sirf speed aur wavelength shrink hote hain Sun ka blackbody peak wavelength aur band? ::: 500 nm, visible