I found an answer from en.wikipedia.org

Potassium-40 - Wikipedia

Potassium-40 (⁴⁰K) is a radioactive isotope of potassium which has a long half- life of 1.251×10⁹ years. It makes up 0.012% (120 ppm) of the total amount of ...

For more information, see Potassium-40 - Wikipedia

I found an answer from www.toppr.com

Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A

In this topic, we will learn about the Laws of Radioactive Decay. ... Now, integrating both the sides of the above equation, we get, ... This rate gives us the number of nuclei decaying per unit time. ... Next, let's find the relation between the mean life τ and the disintegration constant λ. ... Download NCERT Notes and Solutions.

For more information, see Law of Radioactive Decay, Decay Rate, Half-Mean Life, Q&A

I found an answer from www.quora.com

What percentage of radioactive isotopes are left at each half-life ...

After 1 half-life there will be 50% of the original isotope, and 50% of the decay product. ... Answered 4 years ago · Author has 5.6K answers and 3.6M answer views ... In the half life of a radioactive isotope, why does only half of the nuclei decay ... undergo radioactive decay, decay until they finally become a stable isotope of ...

For more information, see What percentage of radioactive isotopes are left at each half-life ...

Required formulas:

Radioactive Decay Law is a law that regulates the decay of radioactive materials

After period t, the total number of radioactive nuclei left is N.

N = N_0 e^{- \lambda t}, where, N _0 - original amount of the substance,   \lambda - decay constant and t - time

Half life of the radioactive substance T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

Given that

Half - life of the radio active isotope T_{\frac{1}{2}} = T years

(a) Activity reduced to 3.125%

Step 1: Get an expression for time period in terms of its decay constant

Number of radioactive nuclei left after decay N = 3.125  % of N_0

\frac{N}{N_0} = 3.125 % = \frac{3.125}{100}

We know that N=N_0e^{-\lambda t}\ \Rightarrow\ \frac{N}{N_0}\ =\ e^{-\lambda t}

\frac{3.125}{100} = e^{- \lambda t}

\frac{1}{32} = e^{- \lambda t}

Applying the log on both sides

\log 1 - \log 32 = - \lambda t

- 3.47 = - \lambda t

t = \frac{3.47}{\lambda} ---------------------(1)

Step 2: Calculating the time required to decay

Half life T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

\lambda=\frac{\ln2}{T}

Substituting \lambda value in equation (1)

t = \frac{3.47}{\frac{\ln 2}{T}}

t = \frac{3.47}{0.693} T

t = 5T

Hence,  time required to decay 3.125%,   t = 5T yeras

Similarly,

(b) Activity reduced to 1%

Step 1: expression for time period in terms of its decay constant

\frac{N}{N_0} = \frac{1}{100}

\frac{1}{100} = e^{-\lambda t}

t = \frac{4.61}{\lambda}

Step 2: Calculating the time required to decay

t = \frac{4.61}{\frac{\ln 2}{T}}

t = 6.65 T years

Hence,  time required to decay 1%, t = 6.65 T years