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zhuklara [117]
3 years ago
11

The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.

Physics
1 answer:
Rainbow [258]3 years ago
8 0

Answer: 0.0353\ s^{-1}

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in 2.59\times 10\ s

Sample at any time is given by

N=N_oe^{-\lambda t}

where, \lambda=\text{decay constant}

Put values

\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10

Taking natural logarithm both side

\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}

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