Mass-energy equivalence E² = (pc)² + (mc²)²: Worked examples

2.3.32 · D3 · Physics › Modern Physics › Mass-energy equivalence E² = (pc)² + (mc²)²

Yeh practice arena hai parent note Mass-energy equivalence ke liye. Wahan humne master formula derive kiya tha. Yahan hum use har us tarah ke input par use karte hain jo duniya (aur tumhara exam) throw kar sakti hai.

Definition Charon symbols — 30-second recap

Kisi bhi example se pehle, exactly pin down karo ki har letter ka kya matlab hai (parent note inhe poori tarah build karta hai):

$E$ = ek particle ki total relativistic energy (rest energy + kinetic energy). Hamesha ek non-negative number hota hai, joules (J) mein ya, particle physics mein, electron-volts mein.
$p$ = momentum vector ka magnitude $∣ p ∣$ . Kyunki yeh ek vector ki length hai, $p$ hamesha ek non-negative scalar hota hai ( $p \geq 0$ ). Agar ek particle left ki jagah right move kare, toh sirf $p$ ki direction flip hoti hai — uska magnitude $p$ (aur isliye $p c$ ) unchanged rehta hai. Isliye sirf $p \geq 0$ hi formula mein enter karta hai.
$m$ = rest mass (invariant mass): woh mass jo particle ke apne rest frame mein measure hota hai, har frame mein same. Units: kg.
$c$ = vacuum mein light ki speed, $c \approx 3 \times 1 0^{8}$ m/s. Yeh nature ka ek fixed constant hai; $c$ ya $c^{2}$ se multiply karna hi woh cheez hai jo mass aur momentum ko energy ke same units mein convert karti hai.

Neeche master triangle padho (Figure s01): bottom ke saath coral segment momentum-energy leg $p c$ hai; right par lavender segment fixed rest-energy leg $m c^{2}$ hai; dark slate slanted segment hypotenuse $E$ hai. Jaise tum examples se guzarte ho, picture karo ki in do legs mein se kaun zero tak shrink hoti hai ya dominate karti hai — yeh literally wohi hai jo har "case class" is ek triangle ke saath karta hai.

The scenario matrix

Kuch bhi solve karne se pehle, har distinct situation ki list banao jo yeh formula produce kar sakta hai. Har row ek "case class" hai. Baad ke worked examples mein se har ek ek ya zyada cells hit karta hai, aur milke poora table cover karte hain.

#	Case class	Kya special hai	Kaun sa leg dominate / degenerate karta hai	Example
C0	Fully degenerate ( $m = 0$ aur $p = 0$ )	koi mass nahi, koi motion nahi	dono legs $= 0$ → $E = 0$ (koi particle hi nahi)	Ex 0
C1	Pure rest ( $p = 0$ )	koi motion nahi	$p c$ leg $= 0$ → $E = m c^{2}$	Ex 1
C2	Massless ( $m = 0$ )	photon	$m c^{2}$ leg $= 0$ → $E = p c$	Ex 2
C3	General massive	dono legs nonzero	do real legs wala triangle	Ex 3
C4	Non-relativistic limit ( $p c ≪ m c^{2}$ )	slow particle	chhoti $p c$ leg → $\frac{1}{2} m v^{2}$ recover karo	Ex 4
C5	Ultra-relativistic limit ( $p c ≫ m c^{2}$ )	near light-speed	badi $p c$ leg → $E \approx p c$	Ex 5
C6	Solve backwards (given $E$ , $m$ → find $p$ )	formula invert karo	ek leg ke liye rearrange karo	Ex 6
C7	Real-world word problem	mass defect / binding	$Δ m c^{2}$ release hota hai	Ex 7
C8	Exam twist ( $E, m$ se $v$ nikalo)	$γ$ ke through chain karo	formula + $γ$ combine karo	Ex 8

Recall Quick self-test: kaun sa cell?

Ek neutron rest par decay karta hai. "Decay se pehle neutron ki energy" kaun sa case describe karta hai? ::: C1 — pure rest, $E = m c^{2}$ . Ek gamma ray detector se takraati hai. Kaun sa case? ::: C2 — massless, $E = p c$ . Dono legs zero — kya bachta hai? ::: C0 — $E = 0$ , matlab koi particle nahi.

Example 0 — Cell C0: fully degenerate corner ( $m = 0$ aur $p = 0$ )

Worked example Agar DONO legs vanish ho jayein toh?

Maano koi tumhe yeh impossible-looking case de: ek "cheez" jiska zero rest mass aur zero momentum hai. Master formula ise kitni total energy assign karta hai?

Forecast: Agar vertical leg $m c^{2} = 0$ aur horizontal leg $p c = 0$ hai, toh triangle ek single point tak shrink ho jaata hai. Dono legs zero wale triangle ka hypotenuse zero hota hai, toh $E = 0$ .

Step 1 — Dono legs ko simultaneously collapse karo. $E^{2} = (0)^{2} + (0)^{2} = 0 \Rightarrow E = 0$ Yeh step kyun? Yeh woh single input hai jiske liye positive root bhi zero deta hai — momentum-energy aur rest-energy dono contributions absent hain.

Step 2 — Physically interpret karo. $E = 0$ ka matlab hai koi particle hai hi nahi: ek real photon $c$ par move karta hai aur isliye hamesha $p > 0$ hota hai (toh $p c \neq = 0$ ), aur mass wale kisi bhi particle ka $m c^{2} \neq = 0$ hota hai. Toh $m = 0, p = 0$ corner koi physical object nahi hai — yeh woh empty limit hai jo har possible triangle collapse ki list complete karta hai. Yeh step kyun? Yeh case table close karta hai: C1 ek leg kill karta hai, C2 doosra, C0 dono kill karta hai. Enumerate karne ke liye kuch bacha hi nahi.

Verify: $0 + 0 = 0 \geq 0$ , $E \geq 0$ ke saath consistent hai. Koi real photon ya massive particle is cell mein nahi aa sakta. ✓

Example 1 — Cell C1: pure rest energy

Worked example Proton ki rest energy

Ek proton bilkul still baitha hai ( $p = 0$ ). Uska rest mass $m = 1.673 \times 1 0^{- 27}$ kg hai. Uski energy joules aur MeV mein nikalo. ( $c = 3 \times 1 0^{8}$ m/s use karo.)

Forecast: Koi momentum nahi toh horizontal leg vanish ho jaata hai, toh triangle ek vertical line mein collapse ho jaata hai — pure $E = m c^{2}$ . Kyunki proton roughly 1836 electrons hota hai, expect karo around $938$ MeV.

Step 1 — Momentum leg kill karo. $p = 0$ set karo aur positive root lo: $E^{2} = (0)^{2} + (m c^{2})^{2} \Rightarrow E = + m c^{2}$ Yeh step kyun? Jab $p = 0$ hota hai, $(p c)^{2}$ term exactly zero hota hai; sirf rest-energy leg survive karti hai. Hum positive root lete hain kyunki energy non-negative hai.

Step 2 — Numbers (SI) plug in karo. $E = (1.673 \times 1 0^{- 27}) (3 \times 1 0^{8})^{2} = 1.506 \times 1 0^{- 10} J$ Yeh step kyun? $c = 3 \times 1 0^{8}$ m/s light ki speed hai, toh $c^{2} = 9 \times 1 0^{16}$ m $^{2}$ /s $^{2}$ ; mass se multiply karo.

Step 3 — MeV mein convert karo. $1.602 \times 1 0^{- 13}$ J/MeV se divide karo: $E = \frac{1.506 \times 1 0 ^{- 10}}{1.602 \times 1 0 ^{- 13}} \approx 940 MeV$ Yeh step kyun? $1 MeV = 1.602 \times 1 0^{- 13}$ J; particle physicists hamesha rest energies MeV mein quote karte hain. (Book value $938$ MeV se thoda gap $c$ ko $3 \times 1 0^{8}$ tak round karne ki wajah se hai.)

Verify: $\sim 938$ – $940$ MeV textbook proton rest energy hai. Units: (kg)(m/s) $^{2}$ = kg·m $^{2}$ /s $^{2}$ = J. ✓

Example 2 — Cell C2: massless photon

Worked example Gamma-ray photon ka momentum

Ek gamma photon ki energy $E = 1.0$ MeV hai. Uska koi rest mass nahi hai. Uska momentum $p$ nikalo. ( $c = 3 \times 1 0^{8}$ m/s use karo.)

Forecast: Koi rest leg nahi → hypotenuse barabar momentum leg ke, toh $E = p c$ exactly. Expect karo $p = E / c$ .

Step 1 — Rest leg kill karo. $m = 0$ set karo aur positive root lo: $E^{2} = (p c)^{2} + 0 \Rightarrow E = + p c \Rightarrow p = \frac{E}{c}$ Yeh step kyun? Ek massless particle ka $m c^{2} = 0$ hota hai; triangle ek flat horizontal line mein degenerate ho jaata hai. Kyunki $p \geq 0$ aur $E \geq 0$ , positive root physical hai.

Step 2 — Energy ko joules mein convert karo. $E = 1.0 MeV = 1.6 \times 1 0^{- 13} J$ Yeh step kyun? SI mein momentum ke liye energy joules mein chahiye.

Step 3 — $c$ se divide karo. $p = \frac{1.6 \times 1 0 ^{- 13}}{3 \times 1 0 ^{8}} = 5.33 \times 1 0^{- 22} kg\cdotm/s$ Yeh step kyun? $p = E / c$ ; units J/(m/s) = kg·m/s. ✓ ( $p$ magnitude hai, ek non-negative scalar.)

Verify: Photon Momentum dekho — ek real photon zero mass ke bawajood momentum carry karta hai. Newton ka $p = m v$ yahan wrongly zero deta.

Example 3 — Cell C3: general massive particle (poora triangle)

Worked example Ek fast electron ki total energy aur kinetic energy

Ek electron ( $m c^{2} = 0.511$ MeV) ka momentum $p c = 1.0$ MeV hai. Total energy $E$ aur kinetic energy $K E$ nikalo.

Forecast: Dono legs comparable hain ( $1.0$ vs $0.511$ ), toh yeh ek genuine two-legged triangle hai. Expect karo $E$ thoda $1.1$ MeV se upar.

Step 1 — Triangle draw karo aur Pythagoras use karo. $E = (p c)^{2} + (m c^{2})^{2} = 1. 0^{2} + 0.51 1^{2}$ Yeh step kyun? Dono legs nonzero hain → poora formula use karo, kuch collapse nahi hota. Positive root, hamesha ki tarah energy ke liye.

Step 2 — Compute karo. $E = 1.0 + 0.261 = 1.261 = 1.123 MeV$ Yeh step kyun? Har leg ko square karo, add karo, root lo — literally hypotenuse.

Step 3 — $K E$ ke liye rest energy strip out karo. $K E = E - m c^{2} = 1.123 - 0.511 = 0.612 MeV$ Yeh step kyun? Total energy = rest + kinetic; jo bhi $m c^{2}$ se zyada hai woh kinetic part hai.

Verify: Triangle sanity check: hypotenuse $1.123 >$ har leg ( $1.0$ , $0.511$ ). ✓ Aur $K E > 0$ expected hai ek moving particle ke liye.

Figure s02 ko is example ke saath padho: yeh exactly isi triangle ko scale par draw karta hai — coral base $p c = 1.0$ MeV momentum leg hai, lavender upright $m c^{2} = 0.511$ MeV rest leg hai, aur slate hypotenuse par computed $E = 1.123$ MeV label hai. Mint annotation leftover slice $K E = E - m c^{2} = 0.612$ MeV dikhata hai. Visually notice karo ki hypotenuse base se sirf thodi zyada lambi hai — isliye $E$ ( $1.123$ ) $p c$ ( $1.0$ ) se just upar hai.

Woh tool jis par hum do baar lean karte hain: binomial approximation

Example 4 — Cell C4: non-relativistic limit (slow particle)

Worked example Dikhao ki formula slow proton ke liye Newton recover karta hai

Ek proton ( $m c^{2} = 938$ MeV) ka momentum bahut chhota hai $p c = 9.38$ MeV (toh $p c / m c^{2} = 0.01$ ). $K E$ nikalo aur dikhao ki yeh Newtonian $\frac{1}{2} m v^{2}$ ban jaata hai.

Forecast: Jab momentum leg rest leg se bahut chhoti hoti hai, hypotenuse barely badhti hai. Expect karo $K E \approx \frac{( p c ) ^{2}}{2 m c ^{2}}$ , jo — ek baar slow-speed relation $p = m v$ insert karne par — familiar $\frac{1}{2} m v^{2}$ mein collapse ho jaayega.

Step 1 — Chhoti leg ke liye square root expand karo. $x = p c / m c^{2} = 0.01$ likho, toh $u = x^{2}$ $1 + u$ mein chhoti quantity hai: $E = m c^{2} 1 + x^{2} \approx m c^{2} (1 + \frac{1}{2} x^{2})$ Yeh step kyun? Yeh exactly upar stated binomial tool hai $u = x^{2}$ ke saath; neglected term order $x^{4}$ ka hai, $x = 0.01$ ke liye utterly negligible.

Step 2 — Kinetic energy read off karo. $K E = E - m c^{2} \approx \frac{1}{2} m c^{2} x^{2} = \frac{( p c ) ^{2}}{2 m c ^{2}} = \frac{p ^{2}}{2 m}$ Yeh step kyun? $m c^{2}$ subtract karne par leftover $\frac{1}{2} m c^{2} x^{2}$ bachta hai; $x = p c / m c^{2}$ substitute karne par $c$ 's cancel ho jaate hain aur $\frac{p ^{2}}{2 m}$ milta hai.

Step 3 — Newton tak pahunchne ke liye slow-speed momentum $p = m v$ insert karo. $v ≪ c$ par $γ \to 1$ hota hai (recall $γ = 1/ 1 - v^{2} / c^{2}$ , upar define hai), toh relativistic momentum $p = γ m v$ Newtonian $p = m v$ mein reduce ho jaata hai. Substitute karo: $K E \approx \frac{p ^{2}}{2 m} = \frac{( m v ) ^{2}}{2 m} = \frac{1}{2} m v^{2}$ Yeh step kyun? Yeh Newton tak asli bridge hai: sirf $p$ ko $m v$ se replace karne ke baad hi $\frac{p ^{2}}{2 m}$ literally schoolbook $\frac{1}{2} m v^{2}$ banta hai.

Step 4 — Exact formula ke against number-check karo. $K E_{approx} = \frac{9.3 8 ^{2}}{2 \times 938} = 0.0469 MeV, K E_{exact} = 9.3 8^{2} + 93 8^{2} - 938 = 0.0469 MeV$ Yeh step kyun? Confirm karta hai ki binomial approximation in speeds par kuch bhi miss nahi karti.

Verify: Approximate aur exact $K E$ 4 decimals tak agree karte hain, aur algebra $\frac{1}{2} m v^{2}$ mein reduce ho jaata hai. ✓ Newton bas Einstein ka slow-speed corner hai.

Example 5 — Cell C5: ultra-relativistic limit (near light-speed)

Worked example Collider par ek 50 GeV electron

Ek electron ( $m c^{2} = 0.511$ MeV $= 0.000511$ GeV) ko accelerate kiya jaata hai toh uska momentum $p c = 50$ GeV hai. $E$ nikalo, aur yeh $p c$ se kitna differ karta hai.

Forecast: Ab momentum leg rest leg se ~100 000 guna badi hai. Triangle almost flat line hai: $E \approx p c$ . Expect karo $E$ barely $50$ GeV se upar.

Step 1 — Poora formula (positive root). $E = + (50)^{2} + (0.000511)^{2} GeV$ Yeh step kyun? Ultra-relativistic case mein bhi hum exact start karte hain, phir dekhte hain chhoti leg drop out hoti hai.

Step 2 — Badi leg factor out karke approximate karo. Kyunki $m c^{2} ≪ p c$ hai, $u = (m c^{2} / p c)^{2}$ (chhota) set karo aur binomial tool phir use karo: $E = p c 1 + u \approx p c (1 + \frac{1}{2} u) \approx 50 + \frac{0.00051 1 ^{2}}{2 \times 50} \approx 50 + 2.6 \times 1 0^{- 9} GeV$ Yeh step kyun? Same binomial approximation, ab $u = (m c^{2} / p c)^{2}$ ke saath; neglected error order $u^{2} \sim 1 0^{- 18}$ ka hai.

Step 3 — Interpret karo. Rest-mass correction $\sim 1 0^{- 9}$ GeV hai — utterly negligible. Yeh step kyun? Colliders par $E \approx p c$ constantly use hota hai; high energy par mass "bhool" jaata hai.

Verify: $E - p c = 2.6 \times 1 0^{- 9}$ GeV $> 0$ (hypotenuse hamesha ek leg se zyada hota hai), aur vanishingly small. ✓

Example 6 — Cell C6: backwards solve (given $E$ aur $m$ , $p$ nikalo)

Worked example Formula invert karo

Ek muon ki rest energy $m c^{2} = 105.7$ MeV aur total energy $E = 200$ MeV hai. Uska momentum $p c$ nikalo.

Forecast: Hum hypotenuse aur ek leg jaante hain; doosri leg solve karo. Expect karo $p c = E^{2} - (m c^{2})^{2}$ , kahin $170$ MeV ke aas paas.

Step 1 — Master formula ko $p c$ leg ke liye rearrange karo. $(p c)^{2} = E^{2} - (m c^{2})^{2} \Rightarrow p c = + E^{2} - (m c^{2})^{2}$ Yeh step kyun? Pythagoras backwards bhi chalta hai: ek leg = $hyp^{2} - doosri leg^{2}$ . Hum positive root lete hain kyunki $p \geq 0$ (yeh magnitude hai).

Step 2 — Plug in karo. $p c = 20 0^{2} - 105. 7^{2} = 40000 - 11172.5 = 28827.5 = 169.8 MeV$ Yeh step kyun? Direct substitution; sab MeV mein rakho toh units cleanly cancel ho jaate hain.

Step 3 — Degeneracy check. Kya $E \geq m c^{2}$ hai? Haan ( $200 > 105.7$ ), toh square root real hai. Yeh step kyun? Agar koi $E < m c^{2}$ deta toh answer imaginary hota — physically impossible, total energy kabhi rest energy se neeche nahi ja sakti.

Verify: Reassemble karo: $169. 8^{2} + 105. 7^{2} = 28832 + 11172 \approx 200$ MeV. ✓

Example 7 — Cell C7: real-world word problem (mass defect)

Worked example Hydrogen ke helium mein fuse hone se release hone wali energy

Jab chaar protons ek helium nucleus mein fuse hote hain, products reactants se $Δ m = 4.8 \times 1 0^{- 29}$ kg halke hote hain ("mass defect"). Har fusion mein kitni energy release hoti hai? ( $c = 3 \times 1 0^{8}$ m/s use karo.)

Forecast: Khoyi hui mass pure-rest form $E = (Δ m) c^{2}$ ke zariye energy mein reappear hoti hai. Chhota mass, lekin $c^{2}$ enormous hai — expect karo kuch MeV.

Step 1 — Case pehchano. Missing mass poori tarah rest par energy mein convert hoti hai (photons + kinetic), toh $E = Δ m c^{2}$ use karo (C1-type, difference par applied). Yeh step kyun? Binding energy = mass defect × $c^{2}$ ; release hone wale total ke liye koi momentum term nahi chahiye.

Step 2 — Joules mein compute karo. $E = (4.8 \times 1 0^{- 29}) (3 \times 1 0^{8})^{2} = 4.32 \times 1 0^{- 12} J$ Yeh step kyun? Same $m c^{2}$ machine jaise Example 1 mein, ab $Δ m$ aur $c$ light ki speed ke saath.

Step 3 — MeV mein convert karo. $E = \frac{4.32 \times 1 0 ^{- 12}}{1.6 \times 1 0 ^{- 13}} = 27 MeV$ Yeh step kyun? Nuclear energies MeV mein quote hoti hain; yeh ~27 MeV per helium hi Sun ko power karta hai.

Verify: Known solar fusion ≈26–27 MeV per helium-4 release karta hai. ✓ Units: kg·(m/s) $^{2}$ = J. ✓

Example 8 — Cell C8: exam twist (speed nikalo)

Worked example Energy se velocity tak

Ek particle ki total energy $E = 2 m c^{2}$ hai (uski rest energy se do guna). Uski speed $v$ ko $c$ ke fraction mein nikalo.

Forecast: $E = γ m c^{2}$ , toh $γ = 2$ . Bada $γ$ matlab high speed; expect karo $v$ kahin around $0.87 c$ .

Step 1 — $E$ ko Lorentz factor se relate karo. Kyunki $E = γ m c^{2}$ aur $E = 2 m c^{2}$ : $γ = 2$ Yeh step kyun? Lorentz factor $γ$ (is page ke top par define hai) total energy ko rest energy se directly link karta hai.

Step 2 — $v$ ke liye $γ$ invert karo. $γ = \frac{1}{1 - v ^{2} / c ^{2}}$ se: $1 - \frac{v ^{2}}{c ^{2}} = \frac{1}{γ ^{2}} = \frac{1}{4} \Rightarrow \frac{v ^{2}}{c ^{2}} = \frac{3}{4}$ Yeh step kyun? $γ$ ki definition ko algebraically $v / c$ ke liye solve karo.

Step 3 — Root lo. $\frac{v}{c} = \frac{3}{4} = 0.866 \Rightarrow v \approx 0.866 c$ Yeh step kyun? $v / c$ 0 aur 1 ke beech hai, jaise ek massive particle ke liye required hai — yeh kabhi $c$ (light ki speed) tak nahi pahunchta.

Momentum leg se cross-check: $p c = E^{2} - (m c^{2})^{2} = 4 - 1 m c^{2} = 3 m c^{2}$ , aur $p = γ m v$ deta hai $p c = 2 \cdot 0.866 m c^{2} = 1.732 m c^{2} = 3 m c^{2}$ . ✓

Verify: $γ = 2$ , $v = 0.866 c$ , aur $p c$ tak dono routes $3 m c^{2}$ par agree karte hain. ✓

Matrix wrap up karo

Recall Kya humne har cell cover ki?

C0 fully degenerate? ::: Ex 0 (dono legs zero → E = 0, koi particle nahi). C1 rest? ::: Ex 1 (proton rest energy). C2 massless? ::: Ex 2 (photon momentum). C3 general? ::: Ex 3 (fast electron). C4 non-relativistic limit? ::: Ex 4 (slow proton → Newton). C5 ultra-relativistic limit? ::: Ex 5 (50 GeV electron). C6 solve backwards? ::: Ex 6 (muon momentum). C7 real-world / mass defect? ::: Ex 7 (fusion energy). C8 exam twist (find v)? ::: Ex 8 (E = 2mc² → 0.866c).

Connections

Parent · Mass-energy equivalence
Photon Momentum — Example 2 ka $p = E / c$
Relativistic Kinetic Energy — Examples 3, 4
Lorentz Factor — Example 8 ka $γ$
Nuclear Binding Energy — Example 7 ka mass defect
Special Relativity — in sab ke peeche ka framework
Four-Momentum — deeper 4-vector view