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Question: A sample pitchblende is found to contain \(50\% \) uranium \(\left( {{U^{238}}} \right)\) and \(2.42...

A sample pitchblende is found to contain 50%50\% uranium (U238)\left( {{U^{238}}} \right) and 2.425%2.425\% lead. Of this lead only 93%93\% was Pb206P{b^{206}} isotope. If the disintegration constant is 1.52×1010yr11.52 \times {10^{ - 10}}y{r^{ - 1}}. How old the pitchblende deposit.
A)1.65×108yrs1.65 \times {10^8}yrs
B) 3.3×108yrs3.3 \times {10^8}yrs
C) 6.6×108yrs6.6 \times {10^8}yrs
D) 3.3×109yrs3.3 \times {10^9}yrs

Explanation

Solution

We have to know that the radioactive law expresses that the likelihood per unit time that a core will rot is a consistent free of time. This consistency is known as the rot steady and is signified by . This consistent likelihood may change incredibly between various kinds of cores, prompting a wide range of noticed rot rates. The radioactive rot of a certain number of molecules (mass) is outstanding as expected.
N = N0.e - λt{\text{N = }}{{\text{N}}_{\text{0}}}{\text{.}}{{\text{e}}^{{\text{ - }}\lambda {\text{t}}}}
Where N is the number of radioactive nuclei remaining after time t.
N0 is the number of radioactive nuclei in initial time.
The decay constant is λ\lambda .
t is the time required to decay

Complete step by step solution:
We must remember that the radioactivity is a cycle by which the core of a flimsy molecule loses energy by discharging radiation. In a cabinet which was having photographic plates, a modest quantity of Uranium compound was enveloped by a dark paper and was kept in it. Assessment of these plates later came about when there was an introduction. This marvel came to be known as Radioactive Decay. Radioactive components are the components or isotopes which transmit radiation and go through the cycle of radioactivity. In this article, let us find out about radioactive rot law in detail.
Given:
The amount of sample pitch blender is found to contain 50%50\% of uranium which means that the amount of uranium is 50g50\,g.
The percentage of lead present is 2.425%2.425\% .
The percentage of lead isotope is 93%93\% .
The amount of lead present can be calculated as,
The amount of lead present=2.425×93100=2.255 = 2.425 \times \dfrac{{93}}{{100}} = 2.255
The old of the sample pitch blende can be calculated as,
t=2.303λlnNNot = \dfrac{{2.303}}{\lambda }\ln \dfrac{N}{{{N_o}}}
N0=50+2.255=52.255{N_0} = 50 + 2.255 = 52.255
On simplifying we get,
N=50N = 50
Substituting the values in the equation,
t=2.3031.52×1010yr1ln52.25550=3.3×108t = \dfrac{{2.303}}{{1.52 \times {{10}^{ - 10}}y{r^{ - 1}}}}\ln \dfrac{{52.255}}{{50}} = 3.3 \times {10^8}
Therefore, the option B is correct.

Note: We have to know that the atomic particles can't be limited on the grounds that there is no energy because of the presence of a temperamental core in the component's radioisotope. There is a consistent rot in the isotopes to settle itself with an arrival of a great deal of energy as radiations. Change is alluded to as the cycle of isotope changing into a component of a steady core. It can happen both in regular or counterfeit manners.
Half-life period: The half-life is defined as the amount of time it takes for a given isotope to lose half of its radioactivity. We can calculate the half-life period by using the given formula,
t(1/2) = ln2λ{{\text{t}}_{\left( {{\text{1/2}}} \right)}}{\text{ = }}\dfrac{{{\text{ln2}}}}{\lambda }
Where λ\lambda is the decay constant.