A solution containing active cobalt Co 2760 having an activity of 0 .8 μ Ci and decay constant λ is injected in an animal's body. When 1c m ^ 3 blood is drawn from the animal's body after 10h of injection, the activity

Correct answer: C. 5L Let the total volume of blood is v , initial activity A _ 0 = 0 . Qμ C _ i its activity at time t , A = A _ o e^ - , the activity of x volume, A^ 1 = ((A)/(V))X = X ( A _ o V ) e^ - ⇒ V = X ((A0)/(A1)) e^ - ⇒ V = (1 cm ^ 3 )( 8 × 10 ^ - 7 × 3.7 × 10 ^ 10 30060 )(0.84) = 4.97 × 10 ^ 3 cm ^ 2 =4.97liter = 5L . &

JEE Main 201815 April

Exam: JEE Main
Year: 2018
Date: 15 April
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2018 · 15 April · Physics · Nuclear Physics

A solution containing active cobalt Co 2760 having an activity of 0...

A solution containing active cobalt

Co 2760

having an activity of

0 .8 \mu Ci

and decay constant

\lambda

is injected in an animal's body. When

1c{m}^{3}

blood is drawn from the animal's body after

10h

of injection, the activity was found to be

300dpm

(decays per minute), then the total volume of blood in the animal's body is close to

[ 1Ci = 3 .7 \times {10}^{10}dps

(decays per second) and at

t = 10h

the value of

{e}^{- \lambdat} = 0.84 ]

Options

A $4L$
B $6L$
C $5L$
D $7L$

Correct answer

C. $5L$

Step-by-step solution

Let the total volume of blood is

\text{v}

, initial activity

{A}_{0} = 0 . Q\mu{C}_{i}

its activity at time

\text{t}

A = {A}_{o} e^{- \lambdat}

, the activity of

\text{x}

volume,

A^{1} = \left(\dfrac{A}{V}\right)X = X

\left(\dfrac{{A}_{o}}{V}\right) e^{- \lambdat}

<br />

⇒ V = X \left(\dfrac{A0}{A1}\right)

e^{- \lambdat}⇒ V = \left(1{cm}^{3}\right)\left(\dfrac{8 \times {10}^{- 7} \times 3.7 \times {10}^{10}}{30060}\right)\left(0.84\right)

= 4.97 \times {10}^{3}{cm}^{2} =4.97liter = 5L

.
&

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