Decay law — N = N₀ e^(−λt), half-life, activity: Question bank

2.3.22 · D5 · Physics › Modern Physics › Decay law — N = N₀ e^(−λt), half-life, activity

Neeche ke har claim ki neenv char facts par hai jo tumne pehle se earn ki hain:

Decay constant $λ$ = ek nucleus ki constant probability ki woh per unit time decay karega (units $s^{- 1}$ ).
The law: $N = N_{0} e^{- λ t}$ , jahan $N$ time $t$ ke baad bache hue undecayed nuclei ki sankhya hai.
Activity $A = λ N$ = per second count hone wale decays ki sankhya (jo ek detector read karta hai).
Mean life $τ = 1/ λ$ = ek nucleus ki average lifetime (ek nucleus average mein kitna time chalта hai). Ye half-life nahi hai — hum unhe neeche compare karenge.

Figure 1 — master decay curve (half-life aur mean life marked ke saath):

Figure — Decay law — N = N₀ e^(−λt), half-life, activity

Recall Poora step-by-step:

$N = N_{0} e^{- λ t}$ $d N / d t = - λ N$ se kaise nikalta hai Neeche kuch traps kehte hain "exponential maths se nikal ke aata hai." Yahan woh maths hai, seedha spelled out, taaki kisi ko bhi faith par lena na pade. Step 1 — badlaav ka rule. Ek chote se time $d t$ mein, $N$ nuclei mein se har ek ke decay hone ki probability $λ d t$ hai, isliye $λ N d t$ gaayab ho jaate hain. Kyunki $N$ girta hai, change $d N$ negative hai: $\frac{d N}{d t} = - λ N$ Step 2 — ek jaisi cheezein ek saath rakho. $N$ se divide karo aur $d t$ se multiply karo taaki har $N$ baayein baithe, har $t$ daayein — ab har side ko apne aap mein suma ja sake: $\frac{d N}{N} = - λ d t$ Step 3 — saare chote pieces jodo (integrate karo) shuruat se ( $t = 0, N = N_{0}$ ) ab tak ( $t, N$ ): $\int_{N_{0}}^{N} \frac{d N}{N} = - λ \int_{0}^{t} d t \Rightarrow ln N - ln N_{0} = - λ t$ $ln$ yahan kyun aata hai? Hum $1/ N$ quantity ko sum kar rahe hain jab $N$ shrink karta hai. Socho curve $y = 1/ N$ ke neeche patli rectangles stack kar rahe ho (Figure 2): har rectangle ki height $1/ N$ hai aur width $d N$ hai, isliye uska area $d N / N$ hai. $N_{0}$ se neeche $N$ tak ka total shaded area exactly wahi hai jo $ln$ measure karta hai — $ln$ defined hi hai $1/ N$ ke neeche running area ke roop mein. Yahi ek wajah hai logarithm aata hai; yeh $1/ N$ shape ke liye natural "area-adder" hai. Step 4 — $e$ se log undo karo. $ln (N / N_{0}) = - λ t$ likhо, phir dono sides par $e$ lagao ( $e$ cancel karta hai $ln$ ko): $\frac{N}{N _{0}} = e^{- λ t} \Rightarrow N = N_{0} e^{- λ t}$ Yahi poora derivation hai — Figure 1 mein plot dekho. $ln$ aur $e$ ke "kyun" ki gehrai ke liye, dekho Exponential functions and natural log.

Figure 2 — $ln$ kyun aata hai: $y = 1/ N$ ke neeche ka area exactly $ln N$ hai:

True or false — justify karo

Ek nucleus "safer" hota hai jitna zyada woh pehle se survive kar chuka hai, kyunki ab uska number aa gaya hai.

False. Decay memoryless hota hai — agli instant mein decay hone ki probability

λ d t

waisi hi rehti hai chahe nucleus ek second purana ho ya ek million saal. Nuclei age nahi karte aur wear out nahi hote.

Nuclei ki sankhya double karne se half-life double ho jaati hai.

False. Half-life

T_{1/2} = ln 2/ λ \approx 0.693/ λ

sirf

λ

par depend karta hai, jo isotope ki property hai. Zyada nuclei ka matlab sirf zyada activity hai, par har ek utne hi time mein halva hoga.

Agar activity abhi

A

hai, to ek half-life ke baad activity

A /2

hogi.

True. Kyunki

A = λ N

aur

λ

constant hai, activity directly

N

ke proportional hai, isliye woh exactly wahi

e^{- λ t}

curve follow karta hai aur ek

T_{1/2}

mein halva ho jaata hai.

10 half-lives ke baad sample poora decay ho jaata hai.

False. 10 half-lives ke baad remaining fraction

(1/2)^{10} \approx 1/1024

hai — bahut chhota, par zero nahi. Exponential decay asymptotic hai aur mathematically kabhi zero nahi pohonchta (Figure 3 ki flattening tail dekho).

Do half-lives ke baad ek quarter bachta hai, isliye teen ke baad zero.

False. Half-lives multiply hote hain: teen ke baad

(1/2)^{3} = 1/8

bachta hai. Har half-life jo bhi pichle mein bacha tha use halva karta hai; woh kabhi subtract ho ke zero nahi hote (Figure 3 mein seedhi

1 \to \frac{1}{2} \to \frac{1}{4} \to \frac{1}{8}

dikhai deti hai).

Mean life

τ

half-life ke barabar hoti hai kyunki dono "nuclei kitna jeete hain" describe karte hain.

False.

τ = 1/ λ

(mean lifetime) jabki

T_{1/2} = ln 2/ λ \approx 0.693/ λ

(median lifetime) hai, isliye

τ > T_{1/2}

. Rare long-lived survivors average lifetime ko median se upar kheenchte hain — Figure 1 mein dono marked times dekho.

Bade

λ

wala sample har nucleus ke hisaab se zyada radioactive hota hai.

True. Bada

λ

matlab har nucleus per second higher probability se decay karta hai, isliye same

N

ke liye activity

A = λ N

zyada hogi — aur half-life choti hogi.

Radioactive sample ko garam karne ya compress karne se decay tez ho jaata hai.

False.

λ

nucleus ke andar nuclear forces se set hota hai, ordinary temperature, pressure, ya chemistry practically use chhu nahi sakti. Decay rate ek given isotope ke liye fixed hai.

Galti pakdo

Har item neeche Figure 4 mein bhi "wrong path vs. right path" ke roop mein draw hai — letter ko panel se match karo.

"(A)

N = N_{0} /2

time

t

ke baad, isliye

λ = 2/ t

Galat.

1/2 = e^{- λ t}

λ

ko exponent se bahar kheenchne ke liye logarithm lena hoga:

λ = ln 2/ t = 0.693/ t

, na ki

2/ t

. Exponent ko ordinary algebra se cancel nahi kar sakte — sirf

ln

e

ko undo karta hai.

"(B)

λ t

seconds mein hai kyunki

t

seconds mein hai."

Galat.

λ t

dimensionless hona chahiye — tumhe ise exponentiate karna hai. Agar

λ

s^{- 1}

mein hai to

t

seconds mein hona chahiye taaki units cancel ho jaayein; product khud koi unit nahi rakhta.

"(C) Activity constant hai kyunki decay constant

λ

constant hai."

Galat.

A = λ N

;

λ

fixed hai par

N

girta rehta hai, isliye activity bhi girti hai. Ek constant probability jo ek shrinking population par apply ho rahi hai woh shrinking count deti hai.

"(D)

n

half-lives ke baad remaining fraction

1 - n /2

hai."

Galat. Yeh

(1/2)^{n}

hai. Linear formula

n = 2

par

0

hit karta hai aur usse aage negative ho jaata hai, jo impossible hai; Figure 4 galat straight line ko actual curve ke upar overlay karta hai. Halving multiplicative hai, na ki har baar ek fixed subtraction.

"(E)

T_{1/2} = λ / ln 2

Galat — ulta hai.

T_{1/2} = ln 2/ λ \approx 0.693/ λ

. Physics check karo: bada

λ

(fast decay) chota half-life dega, isliye

λ

denominator mein hona chahiye.

"(F) Ek mean life

τ

ke baad aadhe nuclei ja chuke hain."

Galat. Ek mean life ke baad remaining fraction

e^{- 1} \approx 0.37

hai, isliye lagbhag 63% decay ho chuke hain. Aadha tabhi jaata hai jab ek half-life beet jaaye, jo

τ

se chota hota hai.

"(G) 1 curie ki activity matlab 1 decay per second."

Galat. 1 becquerel (Bq) = 1 decay/s hai; 1 curie

= 3.7 \times 1 0^{10}

Bq. Curie ek bada unit hai jo historically 1 gram radium se juda hua hai.

Why questions

Decay rate

N

ke proportional kyun hota hai,

N^{2}

ya kisi constant ke nahi?

Kyunki har nucleus independently probability

λ d t

se decay karta hai; expected number jo decay karenge woh simply count

N

hai us per-nucleus probability se multiply karke, jo

λ N

deta hai. Nuclei ke beech koi interaction nahi hoti jo

N^{2}

produce kare.

Derivation se ek exponential function kyun nikalta hai, na ki ek straight line?

Kyunki loss ki rate amount present ke proportional hai (

d N / d t = - λ N

). Woh function jiska rate of change khud uske proportional ho exactly exponential hai — yahi uski defining property hai (upar boxed step-by-step dekho).

Hum activity measure kar sakte hain par

N

seedha almost kabhi kyun nahi count kar sakte?

Ek detector events (decays per second) register karta hai jo use pohonchte hain, jo

A = λ N

hai; woh sample mein baithे invisible undecayed nuclei nahi dekh sakta. Par kyunki

A = λ N

hai,

A

measure karne se

N

pata chalta hai.

Ek hi

λ

N

aur

A

dono ko kyun govern karta hai?

Kyunki

A = λ N

bas

N

ko constant

λ

se scale karna hai; kisi curve ko constant se scale karne se uski shape ya half-life nahi badlti, isliye dono

e^{- λ t}

ki tarah decay karte hain.

Mean life half-life se lamba kyun hota hai?

Half-life median time hai (count halva hota hai), par nuclei ki ek choti population bahut lambe time tak survive karti hai. Ye long-lived stragglers mean lifetime ko median time se upar kheenchte hain.

λ = ln 2/ T_{1/2}

use karne se pehle

t

aur

T_{1/2}

ko same units mein convert kyun karna chahiye?

Kyunki

λ

apni units

T_{1/2}

se leta hai, aur phir

λ t

dimensionless hona chahiye. Hours aur seconds mix karna silently exponent ko rescale kar deta hai aur bilkul galat answer deta hai.

Edge cases

Exactly

t = 0

par

N

kya hai?

N = N_{0} e^{0} = N_{0}

— poori starting population, jaisi honi chahiye. Formula correctly us initial condition par anchor karta hai jo integration mein use ki gayi thi.

t \to \infty

hone par

N

ka kya hota hai?

e^{- λ t} \to 0

, isliye

N \to 0

par kabhi actually zero nahi hota. Decay asymptotically axis ke paas jaata hai; mathematically hamesha ek tiny remnant rehta hai (Figure 3 ki tail).

Agar

λ = 0

ho (ek bilkul stable, non-radioactive nuclide)?

Tab

N = N_{0} e^{0} = N_{0}

hoga hamesha — kuch decay nahi hota. Half-life

ln 2/ λ

infinite ho jaata hai, jo correctly kehta hai "kabhi halva nahi hoga."

Agar half-life tumhare measuring time step se bahut choti ho?

Pehla measurement khatam hone se pehle lagbhag sab kuch decay ho jaata hai, isliye tum near-zero activity record karoge aur galti se soch sakte ho ki sample inactive tha. Bahut short-lived isotopes ke liye bahut fast detectors chahiye.

Kya ek single nucleus ke baare mein poochna meaningful hai ki woh kab decay karega?

Nahi — ek nucleus ke liye tum sirf ek probability de sakte ho, kabhi definite time nahi. Smooth exponential law ek statistical statement hai jo sirf nuclei ki badi sankhya ke liye emerge hota hai.

Kya ordinary decay ke dauran activity apni initial value

A_{0}

se zyada ho sakti hai?

Nahi.

A = A_{0} e^{- λ t}

sirf decrease hota hai ek single decaying isotope ke liye. (Apparent rise tabhi hota hai jab decay chains mein daughter feed ho raha ho — ek alag scenario.)

Jab nuclei ki sankhya kuch handful tak gir jaaye, tab bhi smooth exponential curve unhe exactly describe karti hai?

Nahi — sirf kuch nuclei bache hone par, randomness dominate karta hai aur actual count curve ke around integer steps mein jump karta hai. Exponential ek average hai jo sirf bahut bade

N

ke liye exact banta hai.

Connections

Radioactivity — alpha, beta, gamma decay
Exponential functions and natural log
First-order chemical kinetics
Carbon dating

Recall Har trap ki ek-line summary

Randomness memoryless hoti hai, half-lives multiply hote hain, decay asymptotic hota hai (kabhi zero nahi), $λ$ heat/pressure se immutable hai, $τ$ (mean) $> T_{1/2}$ (median), aur activity girती hai chahe $λ$ constant ho kyunki $N$ girta hai.