Question

the radioactivity of a sample is R1 at a time T1 and R2 at time T2. if the half life of the specimen is T,the number of atoms that have disintegrated in the time (T2-T1) is proportional to :
(1) R1T1-R2T2 (2) R1-R2
(3) (R1-R2)/T (4)(R1-R2)T
please explain all the necessary steps in detail

Answer 1

Answer:

          4) $(R_{1} -R_{2} )T$

Explanation:

    Disintegration Constant, λ = $\frac{In 2}{T}$

     Radioactivity, $\frac{dN}{dT} =$ λN

  (Since N(t) = $N_{0}$$e^{-}$λt)

Given :

          $R_{1} =$ λN($T_{1}$)

          $R_{2} =$ λN($T_{2}$)

No. of disintegrated atoms = $N(T_{2}) -N(T_{1} )$ = ${R_{2} -R_{1} }{}$/λ =  $(R_{2}-R_{1}) T$/In2 ]

                                             = $(R_{1} -R_{2} )T$

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Answer 2

The correct option is, (4) (R1-R2)T.

What is radioactive?

• As its name implies, radioactivity is the act of emitting radiation spontaneously.
• This is done by an atomic nucleus that, for some reason, is unstable.

What is an example of radioactive?

• For example, uranium and thorium are two radioactive elements found naturally in the Earth's crust.
• Over billions of years, these two elements slowly change form and produce decay products such as radium and radon.

According to the question:

Radioactivity $=\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$

$\left(\right. Since \left.N(t)=N_{o} e^{-\lambda t}\right)$

Given:

$\begin{aligned}&\mathrm{R}_{1}=\lambda \mathrm{N}\left(\mathrm{T}_{1}\right) \\&\mathrm{R}_{2}=\lambda \mathrm{N}\left(\mathrm{T}_{2}\right)\end{aligned}$

No. of disintegrated atoms $=N\left(T_{2}\right)-N\left(T_{1}\right)$

$=\frac{R_{2}-R_{1}}{\lambda}=\frac{\left(R_{2}-R_{1}\right) T}{\ln 2}$

Hence, the number of atoms that have disintegrated in the time (T2-T1) is proportional to (R1-R2)T.

Learn more about radioactive here,

https://brainly.in/question/5500613?msp_poc_exp=5

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