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1 mg of a radioactive material with half life 1600 years is kept for 2000 years. Choose the correct option.
  • a)
    0.42 mg  material decayed in 2000 years
  • b)
    0.42 mg  material was left after 2000 years 
  • c)
    0.58 mg  material decayed in 2000 years
  • d)
    0.58 mg  material remained after 2000 years
Correct answer is option 'B,C'. Can you explain this answer?
Verified Answer
1mgof a radioactive material with half life 1600 years is kept for 200...


mass remaining after 2000 years is
mass decayed = 0.58 mg
The correct answers are: 0.42 mg  material was left after 2000 years, 0.58 mg  material decayed in 2000 years
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Most Upvoted Answer
1mgof a radioactive material with half life 1600 years is kept for 200...
The radioactive decay equation
The radioactive decay of a material can be described using the radioactive decay equation:
N(t) = N0 * (1/2)^(t/T)
Where:
- N(t) is the amount of radioactive material remaining at time t
- N0 is the initial amount of radioactive material
- t is the time that has elapsed
- T is the half-life of the material

Calculating the amount of material decayed
To calculate the amount of material decayed in 2000 years, we need to subtract the amount of material remaining after 2000 years from the initial amount.

Given:
- Half-life (T) = 1600 years
- Time elapsed (t) = 2000 years

We can plug these values into the radioactive decay equation to find the amount of material remaining:
N(t) = N0 * (1/2)^(t/T)
N(2000) = N0 * (1/2)^(2000/1600)
N(2000) = N0 * (1/2)^(5/4)
N(2000) = 0.7071 * N0

The amount of material decayed is the difference between the initial amount and the amount remaining:
Decayed = N0 - N(2000)

Calculating the amount of material remaining
To calculate the amount of material remaining after 2000 years, we can use the equation N(2000) = 0.7071 * N0, which we derived earlier.

Calculating the amount of material decayed and remaining
Plugging in the values, we find:
Decayed = N0 - N(2000)
Decayed = N0 - 0.7071 * N0
Decayed = 0.2929 * N0

Remaining = N(2000) = 0.7071 * N0

Answer
From the calculations above, we can conclude that:
- Option a) 0.42mg material decayed in 2000 years is incorrect
- Option b) 0.42mg material was left after 2000 years is correct
- Option c) 0.58mg material decayed in 2000 years is correct
- Option d) 0.58mg material remained after 2000 years is incorrect

Therefore, the correct options are b) 0.42mg material was left after 2000 years and c) 0.58mg material decayed in 2000 years.
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1mgof a radioactive material with half life 1600 years is kept for 2000 years. Choose the correct option.a)0.42mg material decayed in 2000 yearsb)0.42mgmaterial was left after 2000 yearsc)0.58mgmaterial decayed in 2000 yearsd)0.58mg material remained after 2000 yearsCorrect answer is option 'B,C'. Can you explain this answer?
Question Description
1mgof a radioactive material with half life 1600 years is kept for 2000 years. Choose the correct option.a)0.42mg material decayed in 2000 yearsb)0.42mgmaterial was left after 2000 yearsc)0.58mgmaterial decayed in 2000 yearsd)0.58mg material remained after 2000 yearsCorrect answer is option 'B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 1mgof a radioactive material with half life 1600 years is kept for 2000 years. Choose the correct option.a)0.42mg material decayed in 2000 yearsb)0.42mgmaterial was left after 2000 yearsc)0.58mgmaterial decayed in 2000 yearsd)0.58mg material remained after 2000 yearsCorrect answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1mgof a radioactive material with half life 1600 years is kept for 2000 years. Choose the correct option.a)0.42mg material decayed in 2000 yearsb)0.42mgmaterial was left after 2000 yearsc)0.58mgmaterial decayed in 2000 yearsd)0.58mg material remained after 2000 yearsCorrect answer is option 'B,C'. Can you explain this answer?.
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