Question

The half-life of radioactive radon is \[3.8\] days. the time at the end of which \[1/{20^{th}}\] of the radon sample will remain undecayed is \[(given\;lo{g_{10}}e = 0.4343)\]
A.\[3.8days\]
B.\[16.5days\]
C.\[33days\].
D.\[76days\]

Answer 1

Hint: The radioactive decay depends on the energy gain of a corresponding process.
This depends upon the kind of radioactive process and structure of the nucleus.
To calculate the decay constant. There is a simple tunneling model, if it is an alpha decay

Complete answer:
Radioactive materials have the ability of emitting particles or electromagnetic photons which are called gamma rays. Hence we use the half-life which represents the time taken by the radioactive components to reduce in number by fifty percent.
The cobalt-\[60\] is known as a famous radioactive source. The time required for half decay of an isotope is known as the half life. The decay is known as the process of turning from one element to another element or the process of turning heavier to the lighter version.
The radiation reduction is directly proportional to the number of radioactive atoms. The universal law of radioactive decay helps to relate the activity and the quantity of the radioactive species.
At some interval of time, the activity may increase or decrease
${t_{\dfrac{1}{2}}} = 3.8day$
Hence we can write the equation of radioactive nuclei is,
$N = {N_0}{e^{ - \lambda t}}$
Where,
$\lambda = \dfrac{{\ln 2}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{\ln 2}}{{3.8}}$
Now substituting in the equation we get,
$\dfrac{{{N_0}}}{{20}} = {N_0}{e^{ - \dfrac{{\ln 2}}{{3.8}}t}}$
Now simplify the equation we get,
$t = 16.5$
It requires a \[16.5\] day

Finally, the correct answer is an option (B).

Note:
The radioactive substance activity is measured in terms of the number of disintegrations.
If the active atoms are more then the number of disintegrations per second is also more.
When the decay constant is larger than the more disintegrations occur in one second.