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What is Half Life Formula?

  • Last Updated : 29 Dec, 2021

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:

t1/2 = 0.693 / λ

Where,

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 = – λN

Here, N is the quantity / amount of the substance at any time t. Therefore, 

dN/N = – λ dt

Integrating both sides, 

∫dN/N = ∫-λ dt

logeN |NoN= – λ t |ot

logeN – logeNo = – λt

loge(N/No) = – λt ⇢ (i)

At half-life, the value of N reduces to half of the initial value. Thus,

N = No / 2

Putting this value in the above equation,

loge(1/2) = -λt1/2

λt1/2 = loge2

t1/2 = loge2 / λ 

Since, loge2 = 0.693, 

t1/2 = 0.693 / λ

Generalized formula:

Nt = No(1/2)t/t1/2

Proof:

From equation (i), 

loge(N/No) = – λt

N/No = e-λt

N = Noe-λt

Since, t1/2 = loge2 / λ, 

λ = log22 / t1/2

Substituting this value, 

N = Noe-loge2 / t1/2× t

N = No(e-loge2 ) t / t1/2

N = No(1/2)t/t1/2

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 

t1/2 = 0.693 / λ = 0.693 / 1.386 = 1/2

t1/2 = 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 t1/2 = 0.2 secs

If the rate constant be λ, then 

t1/2 = 0.693 / λ

λ = 0.693 / t1/2

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 t1/2 = 2λ

Then, write the half-life equation as:

t1/2 = 0.693 / λ

2λ = 0.693 / λ

2 = 0.693

λ2 = 0.3465

λ = √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:

t1/2 = 0.693 / λ

t1/2 = 0.693 / 0.3465 = 2 years

t1/2 = 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 = No(1/2)t/t1/2

Where,

N = 4/4 = 1kg

No = 4kg

t1/2 = 2 years

Putting the values, 

1 = 4 × (1/2)t/2

2t/2 = 4 =22

t/2 = 2

t = 4 years

So, 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 = No(1/2)t/t1/2

Where,

No = 64kgs

t1/2 = 1 year

t = 4 years

Putting the values,

N = 64 × (1/2)4/1 = 64/16 = 4 kgs

Thus, amount of substance decayed = No – N = 64 – 4 = 60kgs

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