# What is Half Life Formula?

The Half-Life formula is used to find the time a substance will decay to half of its initial value. This formula is used where the rate of decay of the substance at any instant is directly proportional to its present value or quantity. As time passes, the quantity of the substance remaining decreases, hence the rate of decay also decreases accordingly. This type of decay is known as First Order Decay.

### Half Life Formula

Half life formula is mostly used for calculating the half-life of radioactive substances which decay proportionally to their present quantity remaining. So, it can be defined **Half-Life** for radioactive substances as the **time taken by one-half of a radioactive substance to decay**. It is important to know about the half-life since knowing about half-lives enables us to determine when a sample of radioactive material is safe to handle or not.

**Formula:**

t_{1/2}= 0.693 / λWhere,

λ = rate constant of the decay

**Proof:**

Let the constant of proportionality of any decay be λ. Then, write the following differential equation:

dN/dt = – λNHere, N is the quantity / amount of the substance at any time t. Therefore,

dN/N = – λ dt

Integrating both sides,

∫dN/N = ∫-λ dt

log

_{e}N |_{No}^{N}= – λ t |_{o}^{t}log

_{e}N – log_{e}N_{o}= – λtlog

_{e}(N/N_{o}) = – λt ⇢ (i)At half-life, the value of N reduces to half of the initial value. Thus,

N = N

_{o }/ 2Putting this value in the above equation,

log

_{e}(1/2) = -λt_{1/2}λt

_{1/2}= log_{e}2t

_{1/2}= log_{e}2 / λSince, log

_{e}2 = 0.693,

t_{1/2}= 0.693 / λ

**Generalized formula:**

N_{t}= N_{o}(1/2)^{t/t}_{1/2}

**Proof:**

From equation (i),

log

_{e}(N/N_{o}) = – λtN/N

_{o}= e^{-λt}N = N

_{o}e^{-λt}Since, t

_{1/2}= log_{e}2 / λ,λ = log

_{2}2 / t_{1/2}Substituting this value,

N = N

_{o}e^{-log}_{e}^{2 / t}_{1/2}^{× t}N = N

_{o}(e^{-log}_{e}^{2 })^{t / t}_{1/2}

N = N_{o}(1/2)^{t/t}_{1/2}

### Sample Problems

**Question 1: Find the value of the half-life of a substance whose decay constant is 1.386 sec ^{-1}.**

**Solution:**

Given, rate constant λ = 1.386

Thus, the value of the half-life is given as

t

_{1/2}= 0.693 / λ = 0.693 / 1.386 = 1/2

t_{1/2}= 0.5 secs

**Question 2: Find the value of the rate constant, given the half-life of a substance, is 0.2 secs.**

**Solution:**

Given half-life t

_{1/2}= 0.2 secsIf the rate constant be λ, then

t

_{1/2}= 0.693 / λλ = 0.693 / t

_{1/2}λ = 0.693 / 0.2 = 3.465 sec

^{-1}

λ = 3.465 sec^{-1}

**Question 3: Given that for a First Order reaction, the half-life is twice the value of the rate constant, find the value of the rate constant of the reaction.**

**Solution:**

Let the rate constant be λ.

Then half-life t

_{1/2}= 2λThen, write the half-life equation as:

t

_{1/2}= 0.693 / λ2λ = 0.693 / λ

2λ

^{2}= 0.693λ

^{2}= 0.3465λ = √0.3465

λ = 0.5886 sec^{-1}

**Question 4: Find the value of the half-life given the value of the rate constant is 0.3465 year ^{-1}.**

**Solution:**

Given the value fo rate constant λ = 0.3465 yr

^{-1}The value of the half-life is given by:

t

_{1/2}= 0.693 / λt

_{1/2}= 0.693 / 0.3465 = 2 years

t_{1/2}= 2 years

**Question 5: Consider a radioactive substance with a mass of 4kg and a** **half-life is 2 years. Find the time when the substance reduces to one-fourth of its present value.**

**Solution:**

Given initial mass = 4kg

Using the generalized formula, we can write

N = N_{o}(1/2)^{t/t}_{1/2}Where,

N = 4/4 = 1kg

N

_{o}= 4kgt

_{1/2}= 2 yearsPutting the values,

1 = 4 × (1/2)

^{t/2}2

^{t/2}= 4 =2^{2}t/2 = 2

t = 4 yearsSo, the time taken by the substance to reduce to one-fourth of its initial value is

4 years.

**Question 6: Find the amount of radioactive substance decayed in 4 years, given that the initial quantity is 64kgs half-life of the substance is 1 year.**

**Solution:**

Given initial mass = 64kgs

The amount of substance remaining is given by the formula,

N = N

_{o}(1/2)^{t/t}_{1/2}Where,

N

_{o}= 64kgst

_{1/2}= 1 yeart = 4 years

Putting the values,

N = 64 × (1/2)

^{4/1}= 64/16 = 4 kgsThus, amount of substance decayed = N

_{o}– N = 64 – 4 =60kgs