EM spectrum — all bands and applications: Worked examples

1.8.35 · D3 · Physics › Electromagnetism › EM spectrum — all bands and applications

Yahan use kiya gaya har symbol parent note mein earn kiya gaya tha. Hum unhe reuse kar rahe hain, redefine nahi kar rahe, lekin hum har step par units re-anchor zaroor karte hain, kyunki unit slips wahan hoti hain jahan marks marte hain.

The scenario matrix

Ek "case" ko do cheezein samjho: kaunsi direction mein compute kar rahe ho (kya tumhe $ν$ pata hai aur $λ$ chahiye? ya $E$ pata hai aur $λ$ chahiye?) aur numbers kis regime mein hain (huge radio wavelengths, tiny gamma wavelengths, degenerate/zero-jaisi limits).

#	Cell (case class)	Kya tricky hai	Covered by
A	$ν \to λ$ , low frequency (radio)	huge $λ$ , engineering meaning	Ex 1
B	$ν \to λ$ , high frequency (gamma)	tiny $λ$ , negative exponents	Ex 2
C	$λ \to E$ , energy in eV (visible)	do conversions chained	Ex 3
D	$E \to λ$ , energy given in eV (X-ray)	formula invert karo, pehle convert karo	Ex 4
E	Ratio / no-numbers comparison	constants cancel, no calculator	Ex 5
F	Limiting / degenerate case ( $ν \to 0$ , $ν \to \infty$ )	edges par kya hota hai	Ex 6
G	Word problem (microwave oven, real device)	physics ko ek number mein translate karo	Ex 7
H	Boundary between bands (ek band kahan khatam hota hai?)	ek threshold wavelength	Ex 8
I	Exam twist — mixed units, wrong-dikhta data	trap pakdo, unit hygiene	Ex 9
J	Doppler-free consistency check — apna band khud verify karo	apna answer sanity-check karo	Ex 10

Constants jo har jagah use hote hain: $c = 3 \times 1 0^{8} m/s, h = 6.63 \times 1 0^{- 34} J\cdots, 1 eV = 1.6 \times 1 0^{- 19} J .$

Figure — EM spectrum — all bands and applications

Upar wali picture map hai. Notice karo inverse tilt: jaise tum right slide karte ho (frequency upar), wavelength neeche slide karti hai. Neeche ka har example bas is line par ek point pick karna aur ek coordinate read karna hai.

Worked examples

Worked example Ex 1 — Cell A: radio,

$ν \to λ$ , low frequency Ek AM station $ν = 900 kHz$ par broadcast karta hai. Uski wavelength nikalo.

Forecast: AM ki frequency FM se kam hoti hai (100 MHz ne parent note mein 3 m diya tha). Kam $ν$ ⇒ zyada lamba $λ$ . Guess: saikdon metres.

$λ = c / ν$ likho. Yeh step kyun? Humein $ν$ pata hai, $λ$ chahiye; $c = ν λ$ ko $λ$ ke liye solve kiya.
Convert karo: $900 kHz = 900 \times 1 0^{3} Hz = 9 \times 1 0^{5} Hz$ . Yeh step kyun? Formula ko SI chahiye (Hz = cycles/s); "kilo" matlab $1 0^{3}$ .
$λ = \frac{3 \times 1 0 ^{8}}{9 \times 1 0 ^{5}} = 333 m$ . Yeh step kyun? Direct division; powers check karo: $1 0^{8} /1 0^{5} = 1 0^{3}$ , aur $3/9 = 0.333$ .

Verify: Wapas multiply karo: $ν λ = (9 \times 1 0^{5}) (333) = 3.0 \times 1 0^{8} = c$ . ✓ Aur $333 m$ sach mein "saikdon metres" hai — forecast se match karta hai. (Isliye AM towers itne bade hote hain aur antennas poora $λ$ nahi ho sakte.)

Worked example Ex 3 — Cell C: visible,

$λ \to E$ , answer in eV Red light ki $λ = 650 nm$ hai. Photon energy eV mein nikalo.

Forecast: Red visible ka low-energy end hai; green $2.25$ eV tha. Guess: $2.25$ eV se thoda kam, roughly $1.9$ eV.

$E = \frac{h c}{λ}$ . Yeh step kyun? Humein $λ$ pata hai, $E$ chahiye; parent chain $E = h ν = h c / λ$ $ν$ compute karne ko skip karta hai.
$λ$ convert karo: $650 nm = 650 \times 1 0^{- 9} m = 6.5 \times 1 0^{- 7} m$ . Yeh step kyun? "nano" matlab $1 0^{- 9}$ ; SI metres zaroori hain.
$E = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{6.5 \times 1 0 ^{- 7}} = \frac{1.989 \times 1 0 ^{- 25}}{6.5 \times 1 0 ^{- 7}} = 3.06 \times 1 0^{- 19} J$ . Yeh step kyun? Pehle numerator ( $6.63 \times 3 = 19.89$ , exponent $- 34 + 8 = - 26$ , so $1.989 \times 1 0^{- 25}$ ), phir divide.
To eV: $E = \frac{3.06 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 1.91 eV$ . Yeh step kyun? Joules ko joules-per-eV se divide karo units change karne ke liye.

Verify: $1.91 eV < 2.25 eV$ (green) ✓ — red sach mein kam energetic hai. Units: $\frac{J\cdots \cdot m/s}{m} = J$ . ✓

Worked example Ex 4 — Cell D: X-ray,

$E \to λ$ , energy given in eV Ek diagnostic X-ray photon ki $E = 25 keV$ hai. $λ$ nikalo.

Forecast: keV visible eV-scale se hazaar guna zyada hai, to $λ$ 500 nm se hazaar guna chhoti honi chahiye — around $0.05 nm$ , atomic spacing ke paas.

Pehle energy ko joules mein convert karo: $E = 25 \times 1 0^{3} eV \times 1.6 \times 1 0^{- 19} J/eV = 4.0 \times 1 0^{- 15} J$ . Yeh step kyun? $h$ aur $c$ SI mein hain; tum eV ko $h c / λ$ mein mix nahi kar sakte.
Formula invert karo: $λ = \frac{h c}{E}$ . Yeh step kyun? Humein $E$ pata hai, $λ$ chahiye; algebraically $E = h c / λ$ ko flip karo.
$λ = \frac{( 6.63 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{4.0 \times 1 0 ^{- 15}} = \frac{1.989 \times 1 0 ^{- 25}}{4.0 \times 1 0 ^{- 15}} = 4.97 \times 1 0^{- 11} m$ . Yeh step kyun? Wahi numerator pehle wala; exponent $- 25 - (- 15) = - 10$ .

Verify: $4.97 \times 1 0^{- 11} m \approx 0.05 nm$ — $0.01$ – $10$ nm X-ray band ke andar. ✓ Forecast se match karta hai. Yahi near-atomic size exactly kyun hai ki X-rays Bragg diffraction karte hain.

Worked example Ex 5 — Cell E: ratio comparison, no calculator

Photon P ki wavelength $λ_{P}$ hai; photon Q ki $λ_{Q} = 4 λ_{P}$ hai. Zyada energy kismein hai, aur kitne factor se?

Forecast: Lambi wavelength ⇒ kam energy (inverse). Q ke paas lamba $λ$ hai, to P jeetega, aur shayad exactly 4 se.

Dono energies likho: $E_{P} = \frac{h c}{λ _{P}}$ , $E_{Q} = \frac{h c}{λ _{Q}}$ . Yeh step kyun? Same formula, symbolic rakhte hain taaki constants cancel ho jaayein.
Ratio lo: $\frac{E _{P}}{E _{Q}} = \frac{h c / λ _{P}}{h c / λ _{Q}} = \frac{λ _{Q}}{λ _{P}}$ . Yeh step kyun? $h$ aur $c$ divide out ho jaate hain — koi numbers zaroori nahi; energy $1/ λ$ ki tarah scale karti hai.
$λ_{Q} = 4 λ_{P}$ substitute karo: $\frac{E _{P}}{E _{Q}} = 4$ . Yeh step kyun? $λ_{P}$ cancel ho jaata hai, sirf factor bachta hai.

Verify: $E_{P} = 4 E_{Q}$ : P 4× zyada energetic hai, aur uski wavelength chhoti hai — "energy $\propto 1/ λ$ " ke consistent hai. ✓ No calculator, no units — ek pure structure argument.

Worked example Ex 6 — Cell F: limiting / degenerate cases

$λ$ aur $E$ ka kya hota hai jab (a) $ν \to 0$ aur (b) $ν \to \infty$ ? Kya $ν = 0$ ek real photon hai?

Forecast: Tiny $ν$ ⇒ giant $λ$ , near-zero energy. Huge $ν$ ⇒ shrinking $λ$ , exploding energy. $ν = 0$ ek "bilkul bhi wave nahi" wala edge hona chahiye.

$λ = c / ν$ se: jab $ν \to 0^{+}$ , $λ = c / ν \to \infty$ . Yeh step kyun? Fixed $c$ ko zero ki taraf shrink ho rahe number se divide karo to blow up hoga. Figure ke sabse left point ko infinity ki taraf slide hote dekho.
$E = h ν$ se: jab $ν \to 0^{+}$ , $E \to 0$ . Yeh step kyun? Energy directly $ν$ ke proportional hai; zero frequency matlab zero energy — koi oscillation nahi, koi photon nahi.
Jab $ν \to \infty$ : $λ = c / ν \to 0$ aur $E = h ν \to \infty$ . Yeh step kyun? Mirror limit — map ka right end infinitely energetic, infinitely short waves ki taraf run karta hai.

Verify (degenerate check): $ν = 0$ set karo: $E = h \cdot 0 = 0$ aur $c = 0 \cdot λ$ ka koi finite $λ$ solution nahi hai — to $ν = 0$ ek physical photon nahi hai, yeh empty limit hai. ✓ Dono limits har finite point par $c = ν λ$ respect karte hain. (Yahi edges hain isliye spectrum ek open-ended continuum hai, bounded list nahi — dekho Maxwell's Equations jo kisi bhi $ν$ ko allow karte hain.)

Worked example Ex 7 — Cell G: word problem (microwave oven)

Ek domestic microwave oven $ν = 2.45 GHz$ par chalta hai. (a) $λ$ nikalo. (b) Ek aise photon ki energy eV mein nikalo, aur comment karo kya yeh ek chemical bond tod sakta hai ( $\sim$ kuch eV).

Forecast: GHz microwave hai, to $λ$ kuch centimetres honi chahiye. Photon energy microscopic hogi — bond se bahut neeche, parent note ki "not a resonance, it's heating" wali story confirm karte hue.

$λ = c / ν = \frac{3 \times 1 0 ^{8}}{2.45 \times 1 0 ^{9}}$ . Yeh step kyun? $ν$ pata hai, $λ$ chahiye; GHz $= 1 0^{9}$ Hz.
$λ = 0.1224 m \approx 12.2 cm$ . Yeh step kyun? $3/2.45 = 1.224$ , exponent $1 0^{8} /1 0^{9} = 1 0^{- 1}$ .
Photon energy: $E = h ν = (6.63 \times 1 0^{- 34}) (2.45 \times 1 0^{9}) = 1.62 \times 1 0^{- 24} J$ . Yeh step kyun? Direct $E = h ν$ ; yahan $λ$ ki zaroorat nahi.
eV mein: $E = \frac{1.62 \times 1 0 ^{- 24}}{1.6 \times 1 0 ^{- 19}} = 1.02 \times 1 0^{- 5} eV$ . Yeh step kyun? "few eV" bond scale ke against compare karne ke liye convert karo.

Verify: $λ \approx 12 cm$ microwave band mein hai ✓. Aur $1 0^{- 5} eV ≪ 1 eV$ — ek single microwave photon roughly $100, 000 \times$ bahut weak hai ek bond todne ke liye. ✓ To heating ek-photon bond-breaking nahi ho sakti; yeh wahi bulk dielectric relaxation honi chahiye jo parent note ne describe ki thi.

Worked example Ex 8 — Cell H: do bands ke beech boundary

Visible band khatam hota hai aur UV shuru hota hai $λ = 400 nm$ ke paas. Kaunsi frequency aur photon energy (eV) is violet/UV boundary ko mark karti hai?

Forecast: Red se chhota, to Ex 3 ke 1.91 eV se zyada energy — around 3 eV, aur $ν$ $1 0^{15}$ Hz ke paas.

$ν = c / λ = \frac{3 \times 1 0 ^{8}}{400 \times 1 0 ^{- 9}} = \frac{3 \times 1 0 ^{8}}{4 \times 1 0 ^{- 7}} = 7.5 \times 1 0^{14} Hz$ . Yeh step kyun? Boundary ek wavelength hai; use map par place karne ke liye frequency mein convert karo.
$E = \frac{h c}{λ} = \frac{1.989 \times 1 0 ^{- 25}}{4 \times 1 0 ^{- 7}} = 4.97 \times 1 0^{- 19} J$ . Yeh step kyun? Wahi $h c$ numerator; raw energy nikalo.
eV mein: $E = \frac{4.97 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} \approx 3.11 eV$ . Yeh step kyun? Atomic-transition scale mein convert karo.

Verify: $3.11 eV > 1.91 eV$ (red, Ex 3) ✓ — violet red se zyada energetic hai, jaisi honi chahiye. $7.5 \times 1 0^{14} Hz \approx 1 0^{15}$ ✓, forecast aur Atomic spectra scales se match karta hai.

Worked example Ex 9 — Cell I: exam twist (mixed units, spot the trap)

"Ek UV lamp $λ = 250$ par emit karta hai. Uske photons $8 \times 1 0^{- 19}$ J carry karte hain. Kya data self-consistent hai?" (Note: wavelength unit deliberately omit ki gayi hai — ek classic exam trap.)

Forecast: UV energy of $8 \times 1 0^{- 19} J = 5 eV$ UV ke liye plausible hai, to missing unit almost certainly nm hai, m ya μm nahi. Test karte hain.

nm assume karo: $λ = 250 \times 1 0^{- 9} m = 2.5 \times 1 0^{- 7} m$ . Yeh step kyun? Exam expect karta hai tum physically sensible unit supply karo; hypothesis test karo.
Predicted energy: $E = \frac{h c}{λ} = \frac{1.989 \times 1 0 ^{- 25}}{2.5 \times 1 0 ^{- 7}} = 7.96 \times 1 0^{- 19} J$ . Yeh step kyun? Compute karo energy kya honi chahiye agar $λ = 250 nm$ ho.
Stated $8 \times 1 0^{- 19} J$ se compare karo: dono $0.5%$ ke andar agree karte hain. Yeh step kyun? Consistency check — match confirm karta hai unit assumption ko.

Verify: $7.96 \times 1 0^{- 19} \approx 8 \times 1 0^{- 19} J$ ✓, aur $8 \times 1 0^{- 19} /1.6 \times 1 0^{- 19} = 5 eV$ , bilkul UV territory mein. To data consistent hai provided $λ$ nanometres mein hai. Trap avoid kiya: agar tumne $λ = 250 m$ plug in kiya hota, to $E \sim 1 0^{- 27} J$ milta (radio!) — ek "UV lamp" ke liye absurd.

Active recall

Recall Which cell does each trick belong to?

Given $ν$ , want $λ$ — which formula? ::: $λ = c / ν$ (rearranged $c = ν λ$ ). Given $E$ in eV, want $λ$ — first move? ::: Convert eV → joules (× $1.6 \times 1 0^{- 19}$ ) before using $h c / λ$ . Fastest way to compare two photon energies without a calculator? ::: Ratio $E_{1} / E_{2} = λ_{2} / λ_{1}$ — constants cancel. What is the always-available sanity check on any $(ν, λ)$ pair? ::: Their product must equal $c = 3 \times 1 0^{8}$ m/s. As $ν \to 0$ , what happens to $λ$ and $E$ ? ::: $λ \to \infty$ , $E \to 0$ — the empty (non-photon) limit. An exam gives $λ = 250$ with no unit for a UV lamp — what unit do you supply? ::: Nanometres (nm), because that yields a UV-scale energy of ~5 eV.