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Radioactive decay is the process whereby a radionuclide is transformed into a nuclide of another element as a result of the emission of radiation from its nucleus. The terms parent nuclide and daughter nuclide are often used in descriptions of radioactive decay processes.

Radioactive Decay is the process in which an unstable nucleus or atom splits or transforms into some other nucleus or atom by emission of ionized particle in order to gain stability. The nucleus is held by nuclear force which balances the attraction gravitational force and the repulsive electric force. But radioactive elements are unstable. They get easily dissociated into other elements with releasing a high amount of energy. They undergo this decay process due to convert into smaller stable nuclei. They do not have enough binding force to hold their nucleus and also they have a greater number of neutrons which makes them unstable.

A parent nuclide is the nuclide that undergoes decay in a radioactive decay process. A daughter nuclide is the nuclide that is produced in a radioactive decay process.

Radioactive Decay Formula is expressed by

The alternate formula for Radioactive Decay Formula is

The differential equation for Radioactive Decay Formula is mathematically expressed as
The half-life t1/2 of an isotope is the time taken for the nuclei of that isotope to decay to half of its original number.

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Solved problems based on radioactive decay are given below.

### Solved Examples

Question 1: If 100.0mg of neptunium -239 (239Np) decays to 73.36mg after 24 hours. Find the value of $\lambda$ in the growth decay formula for t expressed in days?
Solution:

Since A = 73.36, Ao = 100.0 and t = 1(day) we have to use the equation

73.36 = 100 e $\lambda$ (1)

0.7336 = e $\lambda$

$\lambda$ = ln 0.7336

$\lambda$ = -0.309791358

Thus the daily decay rate is approximately 31%

Question 2: Carbon-14 used for archeological dating, has a half-life of 5.730 years. Find the decay rate for carbon-14?
Solution:

If 100mg of carbon-14 (No = 100) is present, we are given that in 5.730 years (t=5.730) there will be 50mg present (N=50). We we use the formula.

50 = 100e $\lambda$ (5.730)
0.5 = e 5.730$\lambda$
5,730 $\lambda$ = ln 0.5
$\lambda$ = $\frac{ln0.5}{5.730}$
$\lambda$ = -0.525

We can use this value for carbon-14 calculations.

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