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A radioactive simple is undergoing decay. At any time t1, its activity is A and another time t2, the activity is A5. What is the average life time for the sample?

Option 1 -

t 2 − t 1 I n 5

Option 2 -

I n ( t 2 + t 1 ) 2

Option 3 -

t 1 − t 2 I n 5

Option 4 -

I n 5 t 2 − t 1

0 3 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago
    Correct Option - 1


    Detailed Solution:

    A=A0e−λt1  [Radio active decay law]

    A5=A0e−λ (t2−t1)

    ⇒ln5=λ (t2−t1)

    ⇒Average  life=1λ= (t2−t1)ln5

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