Question

A laboratory has $ 1.49\mu g $ of pure $ {}_7^{13}N $ , which has a half life of $ 10.00 $ min (600s).
a). How many nuclei are present initially?
b). What is the activity initially?
c). What is the activity after $ 1.0hr $ ?
d). After approximately how long will the activity drop to less than one per second?

Answer 1

Hint: The number of nuclei will be the maximum initially as time passes and the number will decrease. Number of nuclei can be found out by using the avogadro's number. The relationship between the half-life, and the decay constant is given by
 $ {T_{\dfrac{1}{2}}} = 0.693/\lambda $ .

Complete Step By Step Answer:
a) Number of moles of nuclei in $ {}_7^{13}N $ is $ \dfrac{{1.49 \times {{10}^{ - 6}}}}{{13}} = 1.146 \times {10^{ - 7}} $
Number of nuclei is found out by multiplying this value by the Avagadro’s number.
Number of nuclei = $ 1.146x{10^{ - 7}} \times 6.02 \times {10^{23}} $
= $ 6.9 \times {10^{16}} $ nuclei

b) $ {T_{\dfrac{1}{2}}} = 0.693/\lambda $
Rearranging the above equation,
$ \lambda = \dfrac{{0.693}}{{600}} = 1.16 \times {10^{ - 3}}{s^{ - 1}} $
at t=0,
$ {\left( {\dfrac{{\Delta N}}{{\Delta t}}} \right)_0} = \lambda {N_0} $
$ \Rightarrow \left( {1.16 \times {{10}^{ - 3}}} \right)\left( {6.90 \times {{10}^{16}}} \right) = 8 \times {10^{13}}decays/s $

c) To calculate activity after 1 hour,
1 hour = 60 minutes = 6 half life periods
$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^6} = \dfrac{1}{{64}} $
This means that the activity after one hour will be $ \dfrac{1}{{64}}th $ times the initial activity.
Activity after 1 hour = $ \Rightarrow 8 \times {10^{13}} \times \dfrac{1}{{64}} $
 $ = 1.25 \times {10^{12}}decay/\sec $

d) we want to find the time t when $ \dfrac{{\Delta N}}{{\Delta t}} = 1\sec $
here we can use the equation $ {e^{\lambda t}} = \dfrac{{\left( {\Delta N/\Delta t} \right)}}{{{{\left( {\Delta N/\Delta t} \right)}_0}}} $
 $ \Rightarrow \dfrac{1}{{8 \times {{10}^{13}}}} = 1.25 \times {10^{ - 14}} $
 $ \Rightarrow {e^{\lambda t}} = 1.25 \times {10^{ - 14}} $
Taking natural log on both sides,
 $ \lambda t = \ln (1.25 \times {10^{ - 14}}) $
 $ t = \dfrac{{\ln (1.25 \times {{10}^{ - 14}})}}{\lambda } = 2.76 \times {10^4}s $ $ = 7.67h $
It will take 7.67 hours to drop activity below one.

Note:
The process by which an unstable atomic nucleus loses energy through radiation is known as radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration). The term "radioactive" refers to a substance that contains unstable nuclei. Alpha decay ( $ \alpha $ -decay), beta decay ( $ \beta $ -decay), and gamma decay ( $ \gamma $ -decay) are three of the most common types of decay, all of which involve the emission of one or more particles.