QJEEAdv18Ph6653: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph6653 Solution: If the measurement errors in all the independent quantities are …

The law of radioactive decay gives number of undecayed nuclei N at time t as $N = {N_0}{e^{ - \lambda t}}$ , where N₀ is number of nuclei at t = 0 and $\lambda$ is the decay constant. The number of nuclei decayed till time t is given by ${N_d} = {N_0} - N$ . Thus, the decay constant is given by

$\lambda = {1 \over t}\ln {{{N_0}} \over N} = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}}$ .

Since, N₀ = 3000 and t = 1 s are given without any error, we assume that error in $\lambda$ is due to error in N_d only. Hence, we can write above equation as

$\lambda + \Delta \lambda = {1 \over t}\ln {{{N_0}} \over {{N_0} - ({N_d} + \Delta {N_d})}}$

$= {1 \over t}\ln {{{N_0}} \over {({N_0} - {N_d})(1 - \Delta {N_d}/({N_0} - {N_d}))}}$

$= {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}} - {1 \over t}\ln \left( {1 - {{\Delta {N_d}} \over {{N_0} - {N_d}}}} \right)$

$\approx \lambda + {1 \over t}{{\Delta {N_d}} \over {{N_0} - {N_d}}}$ ,

which gives $\Delta \lambda = {{\Delta {N_d}} \over {t({N_0} - {N_d})}}$ . Substitute $\Delta$ N_d = 40, t = 1 s, and N_d = 1000 to get $\Delta$ $\lambda$ = 0.02.