Question

Activity of radioactive element decreased to one-third of original activity $R_{o}$ in $9\, yr$. After further $9\, yr$, its activity will be

• A. $R_{o}$
• B. $ \frac{2}{3}R_{o}$
• C. $ \frac{R_{o}}{9} $
• D. $ \frac{R_{o}}{6} $

Answer 1

The Correct Option is C

Activity $R+R_{0} e^{-\lambda t} \frac{R_{0}}{3}=R_{0} e^{-9 \lambda}$
$\Rightarrow e^{-9 \lambda}=\frac{1}{3}$ ... (i)
After further $9\, y r R=Re^{-\lambda t}=\frac{R_{0}}{3} e^{-9 \lambda}$ ... (ii)
From Eqs. (i) and (ii),
$R=\frac{R_{0}}{9}$

Answer 2

Ans, The pace of breaking down of radioactive molecules at any moment is straightforwardly corresponding to the quantity of radioactive particles present in the example right then and there.

Pace of breaking down ( – dN/dt) ∝ N

– dN/dt = λ N

where λ is the rot consistent.

The number of molecules presents undecayed in the example at any moment N = No e^-λt

where No is the number of molecules at time t = 0 and N is the number of iotas at time t.

Half-existence of a Radioactive Element

The time in which the half number of particles present at first in any example rots, is called the half-life (T) of that radioactive component.

The connection between half-life and deterioration steady is given by

T = log₂e/λ = 0.6931/λ

The action of a radioactive component is equivalent to its pace of deterioration.

Action R = ( – dN/dt)

Action of the example after time t,

R = Ro e ^{- λt}

Its SI unit is Becquerel (Bq)

Its different units are Curie and Rutherford.

1 Curie = 3.7 * 10¹⁰ rot/s

1 Rutherford = 10⁶ rot/s