Question
Question: 1 g radius is reduced by 2.1 mg in 5 years by alpha decay, calculate the half-life period....
1 g radius is reduced by 2.1 mg in 5 years by alpha decay, calculate the half-life period.
Solution
Hint: We can solve the above problem by using the radioactive decay formula as given below:
N=Noe−λt
Where No is the initial quantity of substance.
N quantity remained after time t.
λ is decay constant.
Complete step-by-step answer:
Given data is
The initial weight of radium is = 1g
Weight that is reduced is = 2.1mg
Therefore the final weight of the radius is initial weight reduced weight is 1g−2.1mg
⇒1g−2.1×10−3g \left\\{ \because 1mg={{10}^{-3}}g \right\\}
⇒0.9979g≈0.998g
Therefore the final weight of radium is 0.998 g.
Since, we have
Initial weight N0 = 1g
Final weight N(t) = 0.998 g
The time required to reduce by 2.1 mg is 5 years.
We can find out !!λ!! (decay Constant) by using radioactive decay formula.
N=N0e- !!λ!! t
We can rearrange it to find the value of !!λ!!
⇒(N0N)=e- !!λ!! t
⇒ln(N0N)=− !!λ!! t
⇒−ln(N0N)= !!λ!! t
⇒ln(NN0)= !!λ!! t \left\\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\\}
Decay constant !!λ!! =t1ln(NN0)
⇒ !!λ!! =51ln(0.9981)
Since we have for half-life there is a relation between decay constant and half life period as given below:
⇒t1/2=λ0.693
Where t1/2 is half life period of substance.
On substituting value of decay constant !!λ!! =4.004005×10−4years.
⇒t1/2=4.004005×10−4years−10.693
⇒t1/2=1730.767years
So the time required for the 1 g radium which reduces by 2.1 mg in years will take 1730.767 years for decay to half of its weight.
Note: As we used the relation between the decay constant and half life period which we got from radioactive decay formula.
Let’s consider radioactive decay formula
N=Noe−λt
When t=t1/2 , N=2N0 because for half life period remaining amount of substance is half of initial amount of substance.
⇒2N0=N0e−λt1/2
⇒21=e−λt1/2
We can write it in terms of natural logarithmic function as
⇒ln(21)=−λt1/2
⇒−ln(21)=λt1/2
⇒ln(2)=λt1/2 \left\\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\\}
⇒t1/2=λln(2)
⇒t1/2=λ0.693