Question

a A radioactive sample has a half life of 8.3x10^4 years. Calculate its disintegration constant and time taken for 25% of its activity to disappear.​

Answer 1

Answer:

It will take 38298.5 years for 25% of its activity to disappear.

Explanation:

Given the half life = 8.3x10⁴ years = $t_{\frac{1}{2} }$

We know that λ (disintegration constant) = log 2 /  $t_{\frac{1}{2} }$

=> λ = log 2 / 8.3x10⁴

=> λ = 3.62 × 10⁻⁶

We know that λ = $\frac{2.303}{t}$ log $\frac{N_{0} }{N}$

where N₀ = initial number of atoms

and N = final number of atoms

Here, N₀ = N₀ and N = 25%*N₀ = N₀/4

Let the time required for 25% of its activity to disappear be t

=>  3.62 × 10⁻⁶ = $\frac{2.303}{t}$ * log $\frac{N_{0} }{\frac{N_{0}}{4} }$

=> 3.62 × 10⁻⁶ = $\frac{2.303}{t}$ * log $\frac{4* N_{0} }{N_{0}}}$

=> 3.62 × 10⁻⁶ = $\frac{2.303}{t}$ * log 4

=> 3.62 × 10⁻⁶ = $\frac{2.303}{t}$ * 0.602

=> 6.01 × 10⁻⁵ = $\frac{2.303}{t}$

=> t = 38298.5 years.

Therefore, it will take 38298.5 years for 25% of its activity to disappear.

Answer 2

Answer:

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Explanation: