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Question
Copy and complete the following table for a radioactive element whose half-life is 10 minutes. Assume that you have 30g of this element at t = 0.
Solution
`∵ N = (N_0)/2^n` , n being no. of half-life spent.
t = 20 min means 2 half-life and t = 30 min means 3 half-life
`therefore N_20min = N_0/2^2 = N_0/4 and N_30min = N_0/2^3 = N_0/8`
`therefore m_20min = m_0/4 = (30 gm)/4 = 7.5 gm`
and `m_30min = m_0/9 = (30 gm)/9 = 3.33 gm`
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