Physics Optics and Modern - How to Solve: Radioactivity (IIT JEE Main + Advanced)

How to Solve: Radioactivity (IIT JEE Main + Advanced)

Introduction

"Mastering radioactivity doesn’t just help you solve half-life problems—it unlocks 5-7 marks in IIT JEE (Main + Advanced), including tricky nuclear reaction questions and carbon dating. One wrong unit or misapplied formula, and you lose marks. This guide gives you the exact steps to solve any radioactivity problem in under 2 minutes."

WHAT YOU NEED TO KNOW FIRST

1. Exponential functions – You must know how to work with ( e^{-kt} ) and logarithms.
2. Basic nuclear notation – Understand what ( ^A_Z X ) means (A = mass number, Z = atomic number).
3. Unit conversions – Be comfortable converting between seconds, minutes, years, and disintegrations per second (Bq).

KEY TERMS & FORMULAS

Key Terms

Formulas (MEMORISE THESE)

1. Decay Law (Exponential Decay)
   [ N = N_0 e^{-\lambda t} ]
2. ( N ) = Number of undecayed nuclei at time ( t )
3. ( N_0 ) = Initial number of nuclei
4. ( \lambda ) = Decay constant (unit: ( s^{-1} ))

5. ( t ) = Time elapsed

6. Half-Life Formula
   [ t_{1/2} = \frac{\ln 2}{\lambda} ]

7. ( t_{1/2} ) = Half-life

8. ( \lambda ) = Decay constant

9. Activity Formula
   [ A = \lambda N ]

10. ( A ) = Activity (Bq)
11. ( \lambda ) = Decay constant

12. ( N ) = Number of undecayed nuclei

13. Carbon Dating Formula
    [ t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N} \right) ]

14. ( t ) = Age of sample
15. ( N_0 ) = Initial ( ^{14}C ) activity (modern sample)

16. ( N ) = Current ( ^{14}C ) activity

17. Nuclear Reaction Notation
    [ ^A_Z X \rightarrow ^{A'}_{Z'} Y + \text{emitted particle} ]

18. Alpha decay: ( ^4_2 He ) emitted
19. Beta decay: ( ^0_{-1} e ) (electron) or ( ^0_1 e ) (positron) emitted
20. Gamma decay: ( \gamma ) (photon) emitted

STEP-BY-STEP METHOD

Step 1: Identify the Given & Required

• Read the question carefully.
• Note down:
• Initial number of nuclei (( N_0 )) or initial activity (( A_0 ))
• Time elapsed (( t ))
• Half-life (( t_{1/2} )) or decay constant (( \lambda ))
• What is asked? (Remaining nuclei, activity, age, etc.)

Step 2: Convert Units if Needed

• If ( t ) is in years/minutes, convert to seconds (if ( \lambda ) is in ( s^{-1} )).
• If ( t_{1/2} ) is given, find ( \lambda ) using: [ \lambda = \frac{\ln 2}{t_{1/2}} ]

Step 3: Choose the Right Formula

• If asked for remaining nuclei/activity: Use ( N = N_0 e^{-\lambda t} ) or ( A = A_0 e^{-\lambda t} ).
• If asked for half-life: Use ( t_{1/2} = \frac{\ln 2}{\lambda} ).
• If asked for age (carbon dating): Use ( t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N} \right) ).

Step 4: Plug in Values & Solve

• Substitute known values into the formula.
• Use a calculator for ( e^{-\lambda t} ) or logarithms.
• Round off to 2-3 significant figures (JEE expects this).

Step 5: Check Units & Final Answer

• Ensure the answer has the correct unit (e.g., Bq for activity, years for age).
• If the answer seems unreasonable (e.g., negative time), recheck calculations.

WORKED EXAMPLES

Example 1 – Basic (Half-Life Calculation)

Question: The decay constant of a radioactive sample is ( 0.001 \, s^{-1} ). Find its half-life.

Solution: 1. Given: ( \lambda = 0.001 \, s^{-1} ) 2. Required: ( t_{1/2} ) 3. Formula: ( t_{1/2} = \frac{\ln 2}{\lambda} ) 4. Calculation:
[ t_{1/2} = \frac{0.693}{0.001} = 693 \, s ] 5. Answer: ( 693 \, s )

What we did and why: - We used the direct formula for half-life since ( \lambda ) was given. - ( \ln 2 \approx 0.693 ) is a standard approximation.

Example 2 – Medium (Remaining Nuclei)

Question: A sample has ( 10^{12} ) radioactive nuclei with a half-life of 5 minutes. How many nuclei remain after 15 minutes?

Solution: 1. Given:
- ( N_0 = 10^{12} )
- ( t_{1/2} = 5 \, \text{min} = 300 \, s )
- ( t = 15 \, \text{min} = 900 \, s ) 2. Find ( \lambda ):
[ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{300} = 0.00231 \, s^{-1} ] 3. Use decay law:
[ N = N_0 e^{-\lambda t} = 10^{12} \times e^{-0.00231 \times 900} ]
[ N = 10^{12} \times e^{-2.079} ]
[ N = 10^{12} \times 0.125 = 1.25 \times 10^{11} ] 4. Answer: ( 1.25 \times 10^{11} ) nuclei

What we did and why: - Converted time to seconds for consistency with ( \lambda ). - Used ( e^{-\lambda t} ) to find remaining nuclei. - Noticed that 15 min = 3 half-lives, so ( N = N_0 / 8 ) (alternative method).

Example 3 – Exam-Style (Carbon Dating)

Question: A wooden artifact has ( ^{14}C ) activity of 5 Bq. A modern sample has 20 Bq. If the half-life of ( ^{14}C ) is 5730 years, estimate the age of the artifact.

Solution: 1. Given:
- ( A = 5 \, \text{Bq} )
- ( A_0 = 20 \, \text{Bq} )
- ( t_{1/2} = 5730 \, \text{years} ) 2. Find ( \lambda ):
[ \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4} \, \text{year}^{-1} ] 3. Use carbon dating formula:
[ t = \frac{1}{\lambda} \ln \left( \frac{A_0}{A} \right) ]
[ t = \frac{1}{1.21 \times 10^{-4}} \ln \left( \frac{20}{5} \right) ]
[ t = 8264 \times \ln (4) ]
[ t = 8264 \times 1.386 = 11460 \, \text{years} ] 4. Answer: ( 11460 \, \text{years} )

What we did and why: - Used activity ratio instead of nuclei count (common in carbon dating). - Applied the logarithmic formula for age calculation. - Recognized that ( \ln(4) = 2 \ln(2) ), which is a useful shortcut.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is all you need to remember for radioactivity in JEE: 1. Decay law: ( N = N_0 e^{-\lambda t} ). If you see half-life, use ( \lambda = \frac{\ln 2}{t_{1/2}} ). 2. Activity: ( A = \lambda N ). If given activity, convert to ( N ) first. 3. Carbon dating: ( t = \frac{1}{\lambda} \ln \left( \frac{N_0}{N} \right) ). Remember, ( N_0 ) is the initial amount. 4. Nuclear reactions: Balance A and Z on both sides. Alpha decay loses ( ^4_2 He ), beta decay changes ( Z ) by ±1. 5. Units matter! Convert time to seconds if ( \lambda ) is in ( s^{-1} ). That’s it. Now go solve those problems in under 2 minutes each!