JEE Challenger
More from Units and Measurements
PhyMedAdv 2018 • Paper 1

Error Determination in Radioactive Decay Constant from Experimental Data

Comprehension Passage

When measurement errors in independent variables are known, the uncertainty in any dependent quantity can be estimated using a series expansion truncated at the first power of error. For instance, consider the functional relation z=x/yz = x/y. If uncertainties in xx, yy, and zz are denoted by Δx\Delta x, Δy\Delta y, and Δz\Delta z, then: z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} To the first order in Δy/y\Delta y/y, the expansion gives (1±Δyy)−1≈1∓Δyy\left(1 \pm \frac{\Delta y}{y}\right)^{-1} \approx 1 \mp \frac{\Delta y}{y}. Since relative errors in independent quantities combine additively, the uncertainty in zz is: Δz=z(Δxx+Δyy)\Delta z = z\left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right) This approximation assumes Δx/x≪1\Delta x/x \ll 1 and Δy/y≪1\Delta y/y \ll 1, allowing higher-order terms to be omitted.

In an experiment, the initial count of radioactive nuclei is 30003000. During the first 1.0 s1.0\text{ s}, 1000±401000 \pm 40 nuclei are observed to decay. For ∣x∣≪1|x| \ll 1, approximate ln⁡(1+x)≈x\ln(1 + x) \approx x to the first power in xx. The error Δλ\Delta \lambda in the determination of the decay constant λ\lambda, in s−1\text{s}^{-1}, is

Options

A

0.040.04

B

0.030.03

C

0.020.02

Correct
D

0.010.01

Topics & Concepts

Units and MeasurementsRadioactivity

Step-by-Step Solution

To find the error Δλ\Delta \lambda in the determination of the decay constant λ\lambda, we use the radioactive decay law:

N=N0e−λtN = N_0 e^{-\lambda t}

where:

  • N0=3000N_0 = 3000 is the initial number of nuclei.
  • ΔNd=1000±40\Delta N_d = 1000 \pm 40 is the number of nuclei that decayed in time t=1.0 st = 1.0\text{ s}.
  • N=N0−ΔNdN = N_0 - \Delta N_d is the number of radioactive nuclei remaining at time tt.

The mean value of remaining nuclei is: N=3000−1000=2000N = 3000 - 1000 = 2000

The uncertainty in the remaining nuclei is equal to the uncertainty in the decayed nuclei: ΔN=40\Delta N = 40

Taking the natural logarithm of the decay law: ln⁡(N0N)=λt  ⟹  λ=1tln⁡(N0N)\ln\left(\frac{N_0}{N}\right) = \lambda t \implies \lambda = \frac{1}{t} \ln\left(\frac{N_0}{N}\right)

Considering the measured value with its error, we have: λ±Δλ=1tln⁡(N0N∓ΔN)=1tln⁡[N0N(1∓ΔNN)−1]\lambda \pm \Delta \lambda = \frac{1}{t} \ln\left(\frac{N_0}{N \mp \Delta N}\right) = \frac{1}{t} \ln\left[\frac{N_0}{N}\left(1 \mp \frac{\Delta N}{N}\right)^{-1}\right]

Expanding the logarithm: λ±Δλ=1tln⁡(N0N)−1tln⁡(1∓ΔNN)\lambda \pm \Delta \lambda = \frac{1}{t}\ln\left(\frac{N_0}{N}\right) - \frac{1}{t}\ln\left(1 \mp \frac{\Delta N}{N}\right)

Using the given approximation ln⁡(1+x)≈x\ln(1 + x) \approx x for ∣x∣≪1|x| \ll 1, we get: ln⁡(1∓ΔNN)≈∓ΔNN\ln\left(1 \mp \frac{\Delta N}{N}\right) \approx \mp \frac{\Delta N}{N}

Substituting this back gives: λ±Δλ≈λ±1t(ΔNN)\lambda \pm \Delta \lambda \approx \lambda \pm \frac{1}{t}\left(\frac{\Delta N}{N}\right)

Thus, the uncertainty in the decay constant is: Δλ=ΔNNt\Delta \lambda = \frac{\Delta N}{N t}

Substituting the given values: Δλ=402000×1.0 s=0.02 s−1\Delta \lambda = \frac{40}{2000 \times 1.0\text{ s}} = 0.02\text{ s}^{-1}

Therefore, the correct option is (C).

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