IIT-JEE 2010 · Paper 2 · Physics Q49

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PhysicsInteger+3Question 49 of 57

To determine the half life of a radioactive element, a student plots a graph of $\ln\left|\dfrac{dN(t)}{dt}\right|$ versus $t$ . Here $\dfrac{dN(t)}{dt}$ 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 years, the value of $p$ is

ln|dN/dt| (vertical, 1 to 6) against time in years (horizontal, 2 to 8) on a dotted grid: a straight line falling from about 4.5 at t = 3 years to about 3.0 at t = 6 years (slope −1/2 per year), continuing to about 2.3 near t = 7.5.

Your answer

Answer and worked solution

Official answer: 8

Decay constant from the graph. $\ln|dN/dt|=\ln(\lambda N_0)-\lambda t$ , a straight line of slope $-\lambda$ . From the graph it falls by 1.5 in 3 years: $\lambda=0.5$ per year.

Factor. $\dfrac{N_0}N=e^{\lambda t}=e^{0.5\times4.16}=e^{2.08}\approx8$ .

Answer: 8

Topics:Radioactive decay

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IIT-JEE 2010 · Paper 2, Physics question 49