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Question: A radio nucleus \({P^{32}}\) with a half life 14.3days is produced in a nuclear reactor at the const...

A radio nucleus P32{P^{32}} with a half life 14.3days is produced in a nuclear reactor at the constant rate, q = 2.7×1092.7 \times {10^9}nuclei per second. Period soon after the beginnings of production of that nuclide will its activities be equal to 1.7×1091.7 \times {10^9} dps is:
A.9.5 day
B.7.5 day
C.6 day
D.None of these

Explanation

Solution

Radionuclide is radioactive nuclide. It is defined as an atom that has excess nuclear energy, making it unstable. The excess energy used in three different ways:
1.Emitted from a nucleus as gamma radiation.
2.Transfer to electrons to release it as a conversion electron.
3.And used to create and emit the new particle from the nucleus.

Complete step by step answer:
First we will discuss about the statement given in the question,
Half life time is the time required for a quantity to reduce to half of its initial value and was discovered by Ernest Rutherford in 1907. The term is commonly used in nuclear physics to describe the unstable atom present in radioactive decay and generally to characterize any type of exponential and non-exponential decay.
Half life time (T) = 14.3 days
Q = 2.7×1092.7 \times {10^9} nuclei per second
R = 1.7×1091.7 \times {10^9} dps
The radio nuclide is formed at the constant rate q.
The rate decay dNdt\dfrac{{dN}}{{dt}} = λ.N\lambda .N
The rate of accumulation = dNdt\dfrac{{dN}}{{dt}} =(q−λ.N)\left( {q - \lambda .N} \right) ………(1)
N= total number of nuclei present in sample
λ\lambda = Constant of proportionality
Integrating equation (1)
∫0NdNq−λN\smallint {}_0^N\dfrac{{dN}}{{q - \lambda N}} = ∫0tdt\smallint {}_0^tdt
−1λ[log⁡(q−λN)]0n- \dfrac{1}{\lambda }\left[ {\log \left( {q - \lambda N} \right)} \right]_0^n = t
T= −1λlog⁡[q−λNq]→2.303log⁡10[qq−λN]→2.303×14.3days0.693log⁡2.7×1091.7×109→9.5days - \dfrac{1}{\lambda }\log \left[ {\dfrac{{q - \lambda N}}{q}} \right] \to 2.303\log {}_{10}\left[ {\dfrac{q}{{q - \lambda N}}} \right] \to \dfrac{{2.303 \times 14.3days}}{{0.693}}\log \dfrac{{2.7 \times {{10}^9}}}{{1.7 \times {{10}^9}}} \to 9.5days

So, the correct answer is Option A.

Note: A radionuclide scan is a way of imaging bones, organs and other parts of the body by using a small dose of a radioactive chemical. There are different types of radionuclide chemical. The one used depends on which organ or part of the body is to be scanned to determine the dis functionality of the organs.