Question

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

Calculate the decay constant of the radioisotope.​

Answer 1

Answer:
`0.0353\ s^(-1)`

Step-by-step explanation:

Given

Radioactive material is found to decrease 40% of its original value in
`2.59* 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* 10)\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59* 10`

Taking natural logarithm both side

`\Rightarrow \lambda=(\ln 2.5)/(25.9)\\\\\Rightarrow \lambda =0.0353\ s^(-1)`