The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute $10$ minutes later, the rate becomes $2250 $ disintegrations per minute The approximate decay constant is : (Take $\log _{10} 188=0274$ )

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

To determine the decay constant $ \lambda $ of a radioactive sample, we need to use the formula related to the disintegration rate (activity) of a radioactive substance over time. The decay formula is expressed as:

A = A_0 e^{-\lambda t}

where:

$ A $ is the activity at time $ t $.
$ A_0 $ is the initial activity.
$ \lambda $ is the decay constant.
$ t $ is the time elapsed.

Given:

Initial activity $ A_0 = 4250 $ disintegrations/minute.
Activity after $ t = 10 $ minutes is $ A = 2250 $ disintegrations/minute.

The decay equation can be rearranged using the logarithm properties to find $ \lambda $:

\frac{A}{A_0} = e^{-\lambda t}

Taking natural logarithm on both sides:

\ln{\left(\frac{A}{A_0}\right)} = -\lambda t

So, we have:

-\lambda = \frac{\ln{\left(\frac{A}{A_0}\right)}}{t}

Substitute the given values:

-\lambda = \frac{\ln{\left(\frac{2250}{4250}\right)}}{10}

Simplify the fraction:

- \lambda = \frac{\ln{(0.5294)}}{10}

Using the logarithmic property $\ln{(a)} = \log_{10}(a) \times \ln{10}$ and given $ \log_{10} 188 = 0.274 $, we know:

\log_{10}(0.5294) = \log_{10}(525.67) - 3\end \approx 0.274 - 3 \approx -0.726 \end

Now, approximate:

-\lambda = \frac{-0.726 \times \ln{10}}{10} \approx \frac{-1.673}{10} \approx -0.1673

Finally, $ \lambda \approx 0.063 \min^{-1} $.

Therefore, the approximate decay constant is $0.063 \,\min ^{-1}$.

The correct answer is: $0.063 \,\min ^{-1}$.

Answer 2