Question 577: Atoms & Nuclei

Question

In a radioactive material the activity at time t
1
is R
1
and at a later time t
2
, it is R
2
. If the decay constant of the material is $$\lambda $$, then

Accepted Answer

R
1
= R
0
e
-$$\lambda $$t
1
and R
2
= R
0
e
-$$\lambda $$t
2
$$ 	herefore $$ $${{{R_1}} \over {{R_2}}}$$ = $${{{R_0}{e^{ - \lambda {t_1}}}} \over {{R_0}{e^{ - \lambda {t_2}}}}}$$ = $${{e^{ - \lambda \left( {{t_1} - {t_2}} ight)}}}$$
$$ \Rightarrow $$ R
1
= R
2
e
$$-$$$$\lambda $$(t
1
$$-$$t
2
)

Answer

R
1
= R
2

Answer

R
1
= R
2
e
$$-$$$$\lambda $$(t
1
$$-$$t
2
)

Answer

R
1
= R
2
e
$$\lambda $$(t
1
$$-$$t
2
)

Answer

R
1
= R
2
(t
2
/t
1
).

Atoms & Nuclei

In a radioactive material the activity at time t 1 is R 1 and at a later time t 2 , it is R 2 . If the decay constant of the material is $\lambda $, then

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