{"id":334,"date":"2024-05-11T05:39:40","date_gmt":"2024-05-11T05:39:40","guid":{"rendered":"https:\/\/casestudyhelp.net\/questions\/?p=334"},"modified":"2024-05-11T05:39:40","modified_gmt":"2024-05-11T05:39:40","slug":"what-is-half-life-formula","status":"publish","type":"post","link":"https:\/\/casestudyhelp.net\/questions\/what-is-half-life-formula\/","title":{"rendered":"What is Half-Life Formula?"},"content":{"rendered":"

What is Half-Life Formula?<\/span><\/h2>\n

Answer:<\/span><\/u><\/strong><\/h3>\n

A half-life is the amount of time needed for half of an atomic nuclide in a given sample to react, or the amount of time needed for neutrons to decay into proton, electron, and antineutrino groups.<\/span> This is a mechanism that occurs in many different kinds of nuclear and chemical incidents, such as the decay of radioactive isotopes or nuclei.<\/span><\/p>\n

The Detection of Half-Lives<\/span><\/strong><\/h3>\n

Ernest Rutherford, one of the leading scientists of his day, is credited with discovering half-lives as well as alpha and beta radiation. Alongside scientist Joseph John Thompson, Rutherford conducted complementary experiments that ultimately led to the discovery of electrons, placing him at the forefront of this momentous discovery. Research on radioactivity started when Rutherford realized the possibilities of what he was seeing. The difference between beta and alpha radiation was discovered by him two years later. Since samples of radioactive materials decayed by half in the same period, this led to his discovery of half-lives.<\/span><\/p>\n

Derivation of the Half Life Formula<\/span><\/strong><\/h3>\n

We first begin with the exponential decay law, which is expressed as follows:<\/span><\/p>\n

N(t) = N0e<\/span> \u2212\u03bbt<\/span><\/p>\n

Additionally, one needs to set t = T1\/2 and N (T1\/2) = \u00bd N0<\/span><\/p>\n

N (T1\/2) = 1\/2N0 = N0e \u2013<\/span> \u03bbT1\/2<\/span><\/p>\n

Next, calculate the logarithm by dividing by N0.<\/span><\/p>\n

\u00bd = e<\/span> \u2212\u03bbt<\/span><\/p>\n

Thus, In (1\/2) = \u2212\u03bbT1\/2<\/span><\/p>\n

Now solving for T1\/2,<\/span><\/p>\n

T1\/2 = \u22121\u03bb In (1\/2)<\/span><\/p>\n

The \u201c-1\u201d can be raised to an exponent of the logarithm using the logarithmic rules. Ultimately, this provides<\/span><\/p>\n

T1\/2 = In (2))\/\u03bb<\/span><\/p>\n

The Half-Life of Certain Substances<\/span><\/strong><\/h3>\n

Here are some examples of elements and their associated half-lives, as the half-life formula has been used to determine the half-lives of several isotopes:<\/span><\/p>\n

Silver-94 \u2013 0.42 seconds<\/span><\/p>\n

Neutron \u2013 10.2 min<\/span><\/p>\n

Iodine-131 \u2013 8 days<\/span><\/p>\n

Carbon-14 \u2013 5,730 years<\/span><\/p>\n

Plutonium-239 \u2013 24,100 years<\/span><\/p>\n

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A Few Examples of the Half-Life Formula<\/span><\/strong><\/h3>\n

Question:<\/span><\/b> If we were to calculate a certain radioactive substance\u2019s half-life and use a decay constant of 0.003 per year, the result would be ____.<\/span><\/p>\n

Solution: <\/span><\/b>t1\/2 = ln (2)\/\u03bb<\/span><\/p>\n

t1\/2 = 0.693 \/ 0.003 yearly<\/span><\/p>\n

t1\/2 = 231 years<\/span><\/p>\n

In other words, this indicates that the half-life of the material in question is 231 years and that half of it will take that long to decompose.<\/span><\/p>\n

Question: <\/span><\/b>Determine the decay constant value of a radioactive material with a half-life of 0.04 seconds.<\/span><\/p>\n

Solution: <\/span><\/b>Given that the substance\u2019s half-life is t1\/2 = 0.04<\/span><\/p>\n

To determine a substance\u2019s half-life, apply the half-life formula.<\/span><\/p>\n

t1\/2 = 0.693\/\u03bb<\/span><\/p>\n

\u03bb= 0.693\/0.04<\/span><\/p>\n

= 17.325<\/span><\/p>\n

Thus, 17.325 s-1 is the radioactive substance\u2019s decay constant.<\/span><\/p>\n

Question: <\/span><\/b>If the material contains 90% of the carbon-14 present in living tissue, then calculate the age of the Shroud of Turin.<\/span><\/p>\n

Solution: <\/span><\/b>The original equation\u2019s N(t)\/N0 equals 90%, or 0.90, which can be changed to 0.90= e(\u03bb*t).<\/span><\/p>\n

Since the value of lambda is unknown, we can utilize the first half-life formula, which yields 0.693 over 5730 years since the half-life of carbon-14 is known.<\/span><\/p>\n

Returning to the previous equation, we can see that ln 0.90 = -\u03bbt by calculating the natural logarithm of both sides.<\/span><\/p>\n

We would get t = ln 0.90\/\u03bb by rearranging this new equation to isolate t.<\/span><\/p>\n

Solving the two equations for time t and changing lambda to the appropriate equation<\/span><\/p>\n

t = ln 0.90\/ (0.693\/5730 years)<\/span><\/p>\n

t = 871 years<\/span><\/p>\n

What Are the Applications of Half-Life?<\/h3>\n

Geologists and archaeologists and even physicians use this formula. Some of the notable applications of half-life are:<\/p>\n