Question 264: Atoms & Nuclei

Question

In a radioactive material, fraction of active
material remaining after time t is 9/16. The
fraction that was remaining after t/2 is

Accepted Answer

First order decay
N(t) = N
0
e
-$$\lambda $$t
Given $${{N\left( t \right)} \over {{N_0}}} = {9 \over {16}} = $$ e
-$$\lambda $$t
N(t/2) = N
0
e
-$$\lambda $$(t/2)
$${{N\left( {t/2} \right)} \over {{N_0}}} = \sqrt {{e^{ - \lambda t}}} $$ = $$\sqrt {{9 \over {16}}} $$ = $${3 \over 4}$$
$$ \Rightarrow $$ $$N\left( {t/2} \right) = {3 \over 4}{N_0}$$

Answer

$${3 \over 4}$$

Answer

$${4 \over 5}$$

Answer

$${3 \over 5}$$

Answer

$${7 \over 8}$$

Atoms & Nuclei

In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is

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