A `280 `days old radioactive substance shown an activity of 6000 dps, 100 days later its activity between 3000 dps ,what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `120000 dps`D. 6000 dps`

A `280 `days old radioactive substance shown an activity of 6000 dps,

1.	A `280 `days old radioactive substance shown an activity of 6000 dps, 100 days later its activity between 3000 dps ,what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `120000 dps`D. 6000 dps`
Answer» Correct Answer - b We know that `lambda=(2.303)/(t) log .A_(0)/(A)` where `A_(0)` is the initial activity. A is the activity at time t. `:. lambda=(2.303)/(280)log.(A_(0))/(600) =(2.303)/(420) log.(A_(0))/(3000)` On solving, we get `A_(0) =24000 dps`

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A `280 `days old radioactive substance shown an activity of 6000 dps, 100 days later its activity between 3000 dps ,what was its initial activity ?A. `20000 dps`B. `24000 dps`C. `120000 dps`D. 6000 dps`

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