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Potassium-40 - Wikipedia

Potassium-40 (^{40}K) **is** a **radioactive isotope** of potassium which **has** a **long half**-
**life** of 1.251×10^{9} **years**. It makes up 0.012% (120 ppm) of the total amount of ...

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**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

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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