How to Calculate Half-Life

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For a substance in the process of decomposition, the time it takes for the amount of substance to be reduced by half is called the half-life or half-life. ^{[1] X Source of Research} Originally, this term was used to describe the decay of a radioactive substance such as uranium or plutonium, however, we can use the term for all substances. has an exponential or cyclic decomposition rate. The half-life of all substances can be calculated by the rate of decomposition, a value calculated based on the initial amount of the substance and the amount remaining after a specified period of time.

Table of Contents

Understanding half-life

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About exponential decomposition. The exponential decay process follows the formula

f

(

x

)

=

a

x

,

{displaystyle f(x)=a^{x},}

f(x)=a^{{x}},

in there

|

a

|

<

first.

{displaystyle |a|<1.}

|a|<1.

In other words, when $x$ ${displaystyle x}$ $x$ increase, $f$ $($ $x$ $)$ ${displaystyle f(x)}$ $f(x)$ decreases and gradually approaches zero. This is the correlation used to describe the half-life. Considering the half-life case, we need $a$ $=$ $first$ $2$ $,$ ${displaystyle a={frac {1}{2}},}$ $a={frac {1}{2}},$ , therefore $f$ $($ $x$ $+$ $first$ $)$ $=$ $first$ $2$ $f$ $($ $x$ $)$ $.$ ${displaystyle f(x+1)={frac {1}{2}}f(x).}$ $f(x+1)={frac {1}{2}}f(x).$

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Rewrite the formula as a half-cycle. This half-life equation does not depend on the variable

x

,

{displaystyle x,}

x,

which depends on time

t

.

{displaystyle t.}

t.

I will get $f$ $($ $t$ $)$ $=$ $($ $first$ $2$ $)$ $t$ ${displaystyle f(t)=left({frac {1}{2}}right)^{t}}$ $f(t)=left({frac {1}{2}}right)^{{t}}$
At this point, what we need to do is not simply put the values into the variable, but consider the actual half-life, in this case, a constant.
Then we need to give the half-life $t$ $first$ $/$ $2$ ${displaystyle t_{1/2}}$ $t_{{1/2}}$ into the exponential equation, however, care should be taken when performing this step. In physics, an exponential equation is an isotropic (direction independent) equation. We know that the amount of a substance depends on time, so we need to divide the quantity of the substance by the half-life – which is a constant with units of time – to get an isotropic quantity.
Thus, we see that $t$ $first$ $/$ $2$ ${displaystyle t_{1/2}}$ $t_{{1/2}}$ and $t$ ${displaystyle t}$ $t$ also have the same unit. Therefore, we get the equation given below.
$f$ $($ $t$ $)$ $=$ $($ $first$ $2$ $)$ $t$ $t$ $first$ $/$ $2$ ${displaystyle f(t)=left({frac {1}{2}}right)^{frac {t}{t_{1/2}}}}$ $f(t)=left({frac {1}{2}}right)^{{{frac {t}{t_{{1/2}}}}}}$

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Take into account the amount of the starting substance. Equation

f

(

t

)

{displaystyle f(t)}

f(t)

We are considering is a correlation equation used to determine the percentage of the substance remaining after a period of time compared to the original amount. Just add the original amount

WOMEN

0

{displaystyle N_{0}}

N_{{0}}

In the above equation, we will get the formula for calculating the half-life of a substance.

$WOMEN$ $($ $t$ $)$ $=$ $WOMEN$ $0$ $($ $first$ $2$ $)$ $t$ $t$ $first$ $/$ $2$ ${displaystyle N(t)=N_{0}left({frac {1}{2}}right)^{frac {t}{t_{1/2}}}}$ $N(t)=N_{{0}}left({frac {1}{2}}right)^{{{frac {t}{t_{{1/2}}}}}}$

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Find the half-life. Usually, the above expression includes all the variables we need to determine the half-life. However, if the substance in question is an unknown radioactive substance, it is possible to determine the mass before and after a period of time, but not its half-life. Thus, we can expand the half-life in terms of measurable variables. This is just a way to transform the expression so that you can easily determine the value to look for. The steps of the transformation process are as follows:

Divide both sides of the expression by the original quantity of substance WOMEN0.{displaystyle N_{0}.} $N_{{0}}.$
- $WOMEN$ $($ $t$ $)$ $WOMEN$ $0$ $=$ $($ $first$ $2$ $)$ $t$ $t$ $first$ $/$ $2$ ${displaystyle {frac {N(t)}{N_{0}}}=left({frac {1}{2}}right)^{frac {t}{t_{1/2}}}}$ ${frac {N(t)}{N_{{0}}}}=left({frac {1}{2}}right)^{{{frac {t}{t_{{1/2}}}} }}$
Get the logarithm base first2{displaystyle {frac {1}{2}}} ${frac {1}{2}}$ On both sides of the expression, we get a simpler expression that doesn’t contain the exponential function.
- $log$ $first$ $/$ $2$ $($ $WOMEN$ $($ $t$ $)$ $WOMEN$ $0$ $)$ $=$ $t$ $t$ $first$ $/$ $2$ ${displaystyle log _{1/2}left({frac {N(t)}{N_{0}}}right)={frac {t}{t_{1/2}}}}$ $log _{{1/2}}left({frac {N(t)}{N_{{0}}}}right)={frac {t}{t_{{1/2}}}}$
Multiply both sides of the expression by tfirst/2{displaystyle t_{1/2}} $t_{{1/2}}$ , then divide both sides by the left side, we get the formula to calculate the half-life. The results will be in logarithmic form, which you can reduce to a regular numerical value using a calculator.
- $t$ $first$ $/$ $2$ $=$ $t$ $log$ $first$ $/$ $2$ $($ $WOMEN$ $($ $t$ $)$ $WOMEN$ $0$ $)$ ${displaystyle t_{1/2}={frac {t}{log _{1/2}left({frac {N(t)}{N_{0}}}right)}}}$ $t_{{1/2}}={frac {t}{log _{{1/2}}left({frac {N(t)}{N_{{0}}}right)}}$

For example

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Example 1. Within 180 seconds, an unidentified radioactive substance decays from its initial mass of 300 g to 112 g. What is the half-life of this substance?

Solution : We have an initial quantity of substance $WOMEN$ $0$ $=$ $300$ $g$ $,$ ${displaystyle N_{0}=300{rm { g}},}$ $N_{{0}}=300{{rm { g}}},$ The remaining amount is $WOMEN$ $=$ $112$ $g$ $.$ ${displaystyle N=112{rm { g}}.}$ $N=112{{rm { g}}}.$ decomposition time is $t$ $=$ $180$ $S$ ${displaystyle t=180{rm { s}}}$ $t=180{{rm { s}}}$ .
The formula for calculating the half-life after transformation is tfirst/2=tlogfirst/2⁡(WOMEN(t)WOMEN0).{displaystyle t_{1/2}={frac {t}{log _{1/2}left({frac {N(t)}{N_{0}}}right)}}.} $t_{{1/2}}={frac {t}{log _{{1/2}}left({frac {N(t)}{N_{{0}}}right)}}.$ . We just need to substitute the values on the right side of the expression and do the calculation to get the half-life of the radioactive substance in question.
- $t$ $first$ $/$ $2$ $=$ $180$ $S$ $log$ $first$ $/$ $2$ $($ $112$ $g$ $300$ $g$ $)$ $\approx$ $127$ $S$ ${displaystyle {begin{aligned}t_{1/2}&={frac {180{rm { s}}}{log _{1/2}left({frac {112{rm { g}}}{300{ rm { g}}}}right)}}&approx 127{rm { s}}end{aligned}}}$ ${begin{aligned}t_{{1/2}}&={frac {180{{rm { s}}}}{log _{{1/2}}left({frac {112{{rm { g}) }}}{300{{rm { g}}}}}right)}}&approx 127{{rm { s}}}end{aligned}}$
Check if the result obtained is reasonable or not. We see that 112 g is less than half of 300 g, so the substance is at least half decayed. Since 127 seconds < 180 seconds, which means that the substance has passed a half-life, the results we obtained here are reasonable.

Advice

There is another way to calculate the half-life using an integer base. In this formula, WOMEN(t){displaystyle N(t)} $N(t)$ and WOMEN0{displaystyle N_{0}} $N_{{0}}$ will reverse the position in the logarithmic function.
- $t$ $first$ $/$ $2$ $=$ $t$ $log$ $2$ $($ $WOMEN$ $0$ $WOMEN$ $($ $t$ $)$ $)$ ${displaystyle t_{1/2}={frac {t}{log _{2}left({frac {N_{0}}{N(t)}}right)}}}$ $t_{{1/2}}={frac {t}{log _{{2}}left({frac {N_{{0}}}{N(t)}}right)}}$
The half-life is a probability-based estimate of the amount of time it takes for a substance to decay to half, not an exact calculation. For example, if there is only one atom of a substance left, it is unlikely that the atom will decay to half an atom after a half-life, but that number of atoms will be zero (zero) or 1 remaining. The larger the residual substance, the more accurate the calculation of the semiconductor period is due to the law of probability for extremely large numbers.

Steps

Understanding half-life

For example

Advice

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