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(a) Plot a graph showing the variation of undecayed nuclei N versus time t. From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.

(b) The air in some caves includes a significant amount of radon gas, which can lead to lung cancer if breathed over a prolonged time. In British caves, the air in the cave with the greatest amount of the gas has an activity per volume of 1.55×105Bq/m3 . Suppose that you spend two full days exploring (and sleeping in) that cave. Approximately how many 222Rn atoms would you take in and out of your lungs during your two-day stay? The radionuclide 222Rn in radon gas has a half-life of 3.82 days. You need to estimate your lung capacity and average breathing rate.Open in App

Solution

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(a) Law of Radioactivity defines that the number of Nuclei undergoing number of Nuclei present in the sample at that Instant.

Since from the graph; we have

N=N−e−λt ....1

where λ=Disintegration constant

For ∴ For T1/2 is the time at N−12ND

⇒ t=T1/2

N02=N0e−λT2

⇒ T1/2=Ln2λ=0.693λ

And for Mean life -we have to sum it over the whole Range for

N(t)=N0e−λt

for number of nuclei which decay in time t to t t △;

N(t)△t=λN0e−λt△t

For Integration it over the Range T=0to∞

τ=λN0∬∞0te−λtdtN0

=λ∫∞0teλtdt

τ=1λ

Mean-life

(b) The equation for the activity is given by:

R=λN

Here, R is the activity, N is the number of nuclei and λ is the decay constant. The equation for the decay constant is given by,

λ=ln2T1/2

Here, T1/2 is the half - life

Thus, R=ln2T1/2N

N=RT1/2ln2

Dividing by volume, NV=RVT1/2ln2

Substituting the value,

=NV=(1.55×105Bq/m3)3.82×24×3600sln2

=7.38×1010 atoms/m3

Thus, these are 7.38×1010 atoms /m3Rn atms per unit volume in the cave.

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