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

2.3.32 · D4 · Physics › Modern Physics › Mass-energy equivalence E² = (pc)² + (mc²)²

Yeh page aapka self-test hai. Problem padho, solution cover karo, khud try karo, phir collapsible kholo aur check karo. Problems paanch levels mein chadhti hain — sirf pieces ko pehchanne se lekar, sab kuch ek hi relativistic scenario mein combine karne tak.

Definition Is page par kaam aane waale symbols aur constants

$c$ = speed of light, space aur time ke beech ka fixed conversion factor. Numerically $c = 3 \times 1 0^{8} m/s$ . Yeh har jagah aata hai kyunki yeh mass ko energy mein convert karta hai ( $m c^{2}$ ) aur momentum ko energy mein ( $p c$ ).
$E$ = total relativistic energy (rest + kinetic).
$p c$ = momentum $p$ ko $c$ se multiply karo, toh energy units mil jaati hain.
$m c^{2}$ = rest energy, invariant rest mass $m$ se bani hai.
$K E$ = kinetic energy, energy ka sirf-motion-wala hissa — jo total se rest energy subtract karne ke baad bachta hai: $K E = E - m c^{2} = (γ - 1) m c^{2}$ . Kam speed par yeh Newtonian $\frac{1}{2} m v^{2}$ ban jaata hai.
$γ$ = Lorentz factor, defined hai $γ = \frac{1}{1 - v ^{2} / c ^{2}}$ . Yeh measure karta hai ki relativity energy aur momentum ko kitna "stretch" karti hai: rest par $γ = 1$ ; jaise $v \to c$ , $γ \to \infty$ . Yeh triangle ko speed se jodta hai $E = γ m c^{2}$ aur $p = γ m v$ ke through. Dekho Lorentz Factor.

Figure — Mass-energy equivalence E² = (pc)² + (mc²)²

Figure s01 — Master triangle. Red hypotenuse $E$ (total energy); bottom black leg $p c$ (motion); right black leg $m c^{2}$ (rest energy, fixed); corner square = right angle. Neeche ke har exercise mein in do sides se ek side find karni hai.

Shuru karne se pehle, ek units reminder jo 90% arithmetic pain door kar deta hai:

Level 1 — Recognition

Recall Solution 1.1

Kaunsa triangle? Horizontal leg $p c = 0$ hai, toh triangle ek vertical line ban jaata hai. Hypotenuse vertical leg ke barabar hoti hai. $E = (0)^{2} + (938)^{2} = 938 MeV$ Yahi famous $E = m c^{2}$ hai — lekin sirf isliye kyunki $p = 0$ . Answer: $E = 938 MeV$ .

Recall Solution 1.2

Kaunsa triangle? Vertical leg $m c^{2} = 0$ hai, toh triangle ek horizontal line ban jaata hai. Hypotenuse horizontal leg ke barabar hoti hai. $E = (2.5)^{2} + (0)^{2} = 2.5 MeV$ Yeh $E = p c$ hai — massless limit. Answer: $E = 2.5 MeV$ . Har photon yahi follow karta hai; dekho Photon Momentum.

Recall Solution 1.3

Kya poochha ja raha hai? Kaun si quantity invariant (frame-independent) hai. $E$ badal jaati hai agar aap particle ke saath saath daudne lago (uski energy alag dikhti hai). $p c$ bhi badal jaata hai (momentum aapki motion par depend karta hai). Lekin $m c^{2}$ — rest energy — rest mass $m$ se bani hai, jo by definition particle ke apne rest frame mein measure hoti hai aur sabke liye same hoti hai. Answer: $m c^{2}$ (rest energy) invariant hai. Deeper reason Four-Momentum mein hai.

Level 2 — Application

Recall Solution 2.1

Step 1 (KYA): Dono legs known hain, toh hypotenuse lo. $E = (500)^{2} + (938)^{2} = 250000 + 880344 = 1130344 \approx 1063.2 MeV$ Step 2 (KYUN subtract karo): Total energy mein rest energy plus motion energy hoti hai. Kinetic energy $K E$ (symbol list mein defined) woh hai jo rest part hatane ke baad bachti hai: $K E = E - m c^{2} = 1063.2 - 938 = 125.2 MeV$ Answer: $E \approx 1063 MeV$ , $K E \approx 125 MeV$ . Dekho Relativistic Kinetic Energy.

Recall Solution 2.2

Kaun si leg missing hai? Hume hypotenuse $E$ aur vertical leg $m c^{2}$ pata hai; hume horizontal leg $p c$ chahiye. Pythagoras rearrange karo: $p c = E^{2} - (m c^{2})^{2} = (1.022)^{2} - (0.511)^{2}$ $= 1.044484 - 0.261121 = 0.783363 \approx 0.885 MeV$ Answer: $p c \approx 0.885 MeV$ . (Neat check: yahaan $E = 2 m c^{2}$ hai, toh is electron ki total energy exactly uski rest energy se double hai.)

Recall Solution 2.3

Step 1 (KYUN $E = p c$ ): Photon massless hai, toh vertical leg vanish ho jaati hai aur $E = p c$ , isliye $p = E / c$ . Step 2 (eV → joules convert karo): $E = 2.0 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 19} J$ . Step 3 ( $c$ se divide karo): $p = \frac{E}{c} = \frac{3.2 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} \approx 1.07 \times 1 0^{- 27} kg\cdotm/s$ Answer: $p \approx 1.07 \times 1 0^{- 27} kg\cdotm/s$ .

Level 3 — Analysis

Recall Solution 3.1

(a) Missing horizontal leg: $p c = E^{2} - (m c^{2})^{2} = 130 0^{2} - 93 8^{2} = 1690000 - 880344 = 809656 \approx 899.8 MeV$ (b) KYUN yeh agla step): Hume speed chahiye, aur momentum-energy aur speed ke beech bridge hai pair $E = γ m c^{2}$ , $p c = γ m v c$ , jahaan $γ = 1/ 1 - v^{2} / c^{2}$ upar definitions box ka Lorentz factor hai. Inhein divide karo taaki $γ$ aur $m$ cancel ho jaayein: $\frac{p c}{E} = \frac{γ m v c}{γ m c ^{2}} = \frac{v}{c}$ Yeh ratio $p c / E$ directly $v / c$ ke barabar hai — ek khoobsurat clean result. Toh: $\frac{v}{c} = \frac{899.8}{1300} \approx 0.692$ Answer: $p c \approx 900 MeV$ , $v \approx 0.692 c$ . Yahaan Lorentz factor hai $\gamma$ $= E / (m c^{2}) = 1300/938 \approx 1.386$ .

Recall Solution 3.2

Hum kya use karein: $m = 0$ ke liye, master relation deta hai $E = p c$ , toh $p c / E = 1$ . Interpretation: Exercise 3.1 mein humne paaya $p c / E = v / c$ . Isko 1 ke barabar karo toh $v / c = 1$ milta hai, yaani $v = c$ . Conclusion: Massless particles zaroori exactly $c$ par travel karte hain — na faster, na slower. Answer: $p c / E = 1 \Rightarrow v = c$ .

Recall Solution 3.3

Step 1 (total energy): $E = K E + m c^{2} = m c^{2} + m c^{2} = 2 m c^{2}$ . Step 2 (momentum): Pythagoras mein plug karo aur horizontal leg isolate karo. $p c = E^{2} - (m c^{2})^{2} = (2 m c^{2})^{2} - (m c^{2})^{2} = 4 (m c^{2})^{2} - (m c^{2})^{2} = 3 m c^{2}$ Answer: $E = 2 m c^{2}$ , $p c = 3 m c^{2} \approx 1.732 m c^{2}$ . (Speed check: $v / c = p c / E = 3 /2 \approx 0.866$ .)

Level 4 — Synthesis

Recall Solution 4.1

Minimum ka MATLAB: Threshold par, do created particles minimum possible energy ke saath produce hote hain — yaani rest mein, har ek sirf apni rest energy lekar. Energy bookkeeping: photon ki energy kam se kam dono rest energies supply karni chahiye: $E_{γ}^{m i n} = m c_{e^{-}}^{2} + m c_{e^{+}}^{2} = 0.511 + 0.511 = 1.022 MeV$ Answer: $E_{γ}^{m i n} = 1.022 MeV$ . (Reality mein ek stray nucleus ko momentum absorb karna padta hai, jo true threshold thoda higher karta hai, lekin energy floor $1.022 MeV$ hai.) Yeh mass–energy conversion in action hai.

Recall Solution 4.2

KYUN energy aati hai: Nuclear reactions mein rest mass conserve nahi hoti; khoyi hui rest mass products ki kinetic energy ke roop mein wapas aati hai. Released energy $Q$ rest energy ki drop hai: $Q = M_{i} c^{2} - M_{f} c^{2} = 3752.0 - 3728.4 = 23.6 MeV$ Answer: $Q = 23.6 MeV$ release hua. "Before" aur "after" triangles ke vertical (rest) legs shrink ho gaye; woh missing rest energy motion energy ban gayi (lambi horizontal legs). Dekho Nuclear Binding Energy.

Recall Solution 4.3

Sign setup: $+ x$ (photon ki direction) ko positive lo. Photon positive $p c$ carry karta hai; atom shuru mein $p c = 0$ se shuru karta hai. Energy conserve karo (KYA): baad mein total energy = atom ki rest energy + photon energy. $E = M c^{2} + E_{γ} = 6000 + 10 = 6010 MeV$ Momentum conserve karo (KYUN alag law): atom rest mein tha ( $p c = 0$ ), photon $p c = + E_{γ} = + 10 MeV$ carry karta tha (massless). Signed momenta add karo, saara kuch atom ko transfer ho jaata hai, jo $+ x$ mein move karta hai: $(p c)_{atom} = + 10 MeV$ Master relation check karo — atom ki nayi rest energy $m^{'} c^{2}$ (sirf $p c$ ka magnitude enter karta hai, kyunki square hota hai): $m^{'} c^{2} = E^{2} - (p c)^{2} = 601 0^{2} - 1 0^{2} = 36120100 - 100 = 36120000 \approx 6009.99 MeV$ Answer: $E = 6010 MeV$ , $p c = + 10 MeV$ , aur excited atom ki rest energy $\approx 6010 MeV$ — absorbed energy almost poori internal (rest) energy mein gayi, thodi si sliver recoil kinetic energy ke roop mein.

Level 5 — Mastery

Recall Solution 5.1

(a) Total energy — energies hamesha positive hoti hain aur simply add hoti hain: $E_{tot} = 511 + 511 = 1022 keV = 1.022 MeV$ (b) Total momentum (YAHAAN SIGNS BITE KARTE HAIN) — $+ x$ positive ke saath: daayein jaata photon $p c = + 511 keV$ hai; bayein jaata photon $p c = - 511 keV$ hai. Hum inhein squaring se pehle unke signs ke saath add karte hain (opposite directions subtract karte hain): $(p_{tot} c) = (+ 511) + (- 511) = 0 keV$ Dhyan raho humne magnitudes nahi add ki (jo galti se $1022$ deta); leftward photon ka negative sign rightward ko cancel kar deta hai. (c) Invariant mass — ab totals square karo: $M c^{2} = E_{tot}^{2} - (p_{tot} c)^{2} = (1022)^{2} - (0)^{2} = 1022 keV = 1.022 MeV$ jo do electron rest energies se match karta hai, $2 \times 0.511 MeV = 1.022 MeV$ — jaisa hona chahiye, kyunki yahi annihilate hua tha. KYUN massless pieces se non-zero: invariant mass individual masses ka sum nahi hai. Yeh total $E$ aur total $p$ se banta hai. Kyunki photons ke signed momenta cancel hokar zero ho gaye, system as a whole ki energy hai lekin net momentum nahi — exactly ek massive object at rest ki tarah. Isliye system aisa behave karta hai jaise uska mass $1.022 MeV / c^{2}$ ho. Answer: $E_{tot} = 1.022 MeV$ , $p_{tot} c = 0$ , $M c^{2} = 1.022 MeV$ . Yeh Four-Momentum ka dil hai.

Recall Solution 5.2

(a) Exact: $E = (50000)^{2} + (938)^{2} = 2.5 \times 1 0^{9} + 879844 \approx 50008.80 MeV$ (b) KYUN yeh approximation: Jab $p c ≫ m c^{2}$ , rest leg momentum leg ke comparison mein tiny hai. $p c$ ko square root se bahar factor karo: $E = p c 1 + (\frac{m c ^{2}}{p c})^{2} \approx p c (1 + \frac{1}{2} (\frac{m c ^{2}}{p c})^{2}) = p c + \frac{( m c ^{2} ) ^{2}}{2 p c}$ binomial expansion $1 + x \approx 1 + \frac{x}{2}$ use karke tiny $x$ ke liye. Numerically: $E \approx 50000 + \frac{93 8 ^{2}}{2 \times 50000} = 50000 + \frac{879844}{100000} = 50000 + 8.798 = 50008.80 MeV$ Fractional error: exact aur approximate values $1$ part in $1 0^{9}$ se bhi better agree karte hain. Answer: $E \approx 50008.80 MeV$ ; approximation error $\sim 1 0^{- 9}$ .

Recall Solution 5.3

(a) Rest energy (invariant leg): $m c^{2} = E^{2} - (p c)^{2} = 250 0^{2} - 240 0^{2} = 6250000 - 5760000 = 490000 = 700 MeV$ (b) Speed golden ratio $v / c = p c / E$ ke through (speed ke liye sirf $p c$ ka magnitude chahiye): $\frac{v}{c} = \frac{2400}{2500} = 0.96$ (c) Lorentz factor $γ = E / (m c^{2})$ ke through: $γ = \frac{2500}{700} \approx 3.571$ Cross-check definition $γ = 1/ 1 - v^{2} / c^{2}$ use karke: $γ = \frac{1}{1 - 0.9 6 ^{2}} = \frac{1}{1 - 0.9216} = \frac{1}{0.0784} = \frac{1}{0.28} \approx 3.571 ✓$ Answer: $m c^{2} = 700 MeV$ , $v = 0.96 c$ , $γ \approx 3.571$ .

Connections

Parent topic
Special Relativity · Lorentz Factor · Photon Momentum
Relativistic Kinetic Energy · Four-Momentum · Nuclear Binding Energy