If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x,y and z are \Delta x,\Delta y and \Delta z, respectively, then z \pm \Delta z = x \pm \Delta x \over y \pm \Delta y = x \over y( 1 \pm \Delta x \over x )( 1 \pm \Delta y \over y )^ - 1. The series expansion for ( 1 \pm \Delta y \over y )^ - 1, to first power in \Delta y/y. is 1 \pm ( \Delta y/y ). The relative errors in independent variables are always added. So the error in z will be \Delta z = z( \Delta x \over x + \Delta y \over y ). The above derivation makes the assumption that \Delta x/x \lt \lt 1,\Delta y/y \lt \lt 1. Therefore, the higher powers of these quantities are neglected. In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000 \pm 40 nuclei decayed in the first 1.0s. For | x | \lt \lt 1.\ln ( 1 + x ) = x up to first power in x. The error \Delta \lambda , in the determination of the decay constant \lambda , in s^ - 1, is

Units and Measurements PYQ — JEE Advanced 2018 · Shift 1 · Q51 with Solution

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Correct answer: C

Use the radioactive decay law

For radioactive decay, $N(t)=N_0 e^{-lambda t}$ where:

initial nuclei: $N_0=3000$
decayed in first $1.0\,s$ : $1000\pm 40$

So the number remaining after $1\,s$ is $N = 3000-1000 = 2000$ with error $\Delta N = 40$ (since $N_0$ is exact and only the decayed count has uncertainty).

Find $\lambda$

Using $2000 = 3000 e^{-\lambda(1)}$ we get $e^{-\lambda} = \frac{2}{3}$ Hence, $\lambda = -\ln\left(\frac{2}{3}\right)=\ln\left(\frac{3}{2}\right)$

Express $\lambda$ in terms of the measured quantity

Let decayed number be $D=1000\pm 40$ . Then remaining number after $1s$ is $N = 3000-D$ And $\lambda = \ln\left(\frac{3000}{N}\right)=\ln\left(\frac{3000}{3000-D}\right)$

We now propagate error from $N$ to $\lambda$ .

Error propagation using logarithm

Since $\lambda = \ln 3000 - \ln N$ only $N$ contributes to error, so $\Delta \lambda = \Delta(\ln N)$ For small error, $\Delta(\ln N) \approx \frac{\Delta N}{N}$ Thus, $\Delta \lambda = \frac{\Delta N}{N} = \frac{40}{2000} = 0.02$

Final answer

$\boxed{\Delta \lambda = 0.02\ \text{s}^{-1}}$

So the correct option is C.

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