To determine the half-life of a radioactive element, a student plots a graph of $\ln \left|\frac{d N(t)}{d t}\right|$ versust. Here $\frac{d N(t)}{d t}$ is the rate of radioactive decay at time $t$ . If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16 \mathrm{yr}$ , the value of $p$ is

Question

To determine the half-life of a radioactive element, a student plots a graph of

\ln \left|\frac{d N(t)}{d t}\right|

versust. Here

\frac{d N(t)}{d t}

is the rate of radioactive decay at time

t

. If the number of radioactive nuclei of this element decreases by a factor of

p

after

4.16 \mathrm{yr}

, the value of

p

is

To determine the half-life of a radioactive element, a student plots a graph of |(d N(t))/(d t)| versust. Here (d

Quantrex Academy · Accepted Answer

Correct answer: A. 8

|(d N)/(d t)|= of radioactive substance| =λ N=λ N_0 e^ -λ t Taking log both sides |(d N)/(d t)|= (λ N_0)-λ t Hence, |(d N)/(d t)| versus t graph is a straight line with slope -λ From the graph we can see that, λ= (1)/(2)= 0.5 yr^ -1 Now applying the equation, aligned N & =N_0 e^ -λ t & =N_0 e^ -0.5 ×4.16 & =N_0 e^ -2.08 =0.125 N_0 & =(N_0)/(8) aligned ie, nuclei decreases by a factor of 8 . Hence the answer is 8 .

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