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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.

Explanation

Solution

Hint: We can solve the above problem by using the radioactive decay formula as given below:
N=NoeλtN={{N}_{o}}{{e}^{-\lambda t}}
Where No{{N}_{o}} is the initial quantity of substance.
NN quantity remained after time t.
λ\lambda 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 1g2.1mg1g-2.1mg
1g2.1×103g\Rightarrow 1g-2.1\times \mathop{10}^{-3}g \left\\{ \because 1mg={{10}^{-3}}g \right\\}
0.9979g0.998g\Rightarrow 0.9979g\approx 0.998g
Therefore the final weight of radium is 0.998 g.

Since, we have
Initial weight N0N_0 = 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!! \text{ }\\!\\!\lambda\\!\\!\text{ } (decay Constant) by using radioactive decay formula.
N=N0e!!λ!! tN=\mathop{N}_{0}\mathop{e}^{\text{- }\\!\\!\lambda\\!\\!\text{ t}}
We can rearrange it to find the value of  !!λ!! \text{ }\\!\\!\lambda\\!\\!\text{ }
(NN0)=e!!λ!! t\Rightarrow \left( \dfrac{N}{\mathop{N}_{0}} \right)=\mathop{e}^{\text{- }\\!\\!\lambda\\!\\!\text{ t}}
ln(NN0)= !!λ!! t\Rightarrow \text{ln}\left( \dfrac{N}{\mathop{N}_{0}} \right)=-\text{ }\\!\\!\lambda\\!\\!\text{ t}
ln(NN0)= !!λ!! t\Rightarrow -\text{ln}\left( \dfrac{N}{\mathop{N}_{0}} \right)=\text{ }\\!\\!\lambda\\!\\!\text{ t}
ln(N0N)= !!λ!! t\Rightarrow \text{ln}\left( \dfrac{\mathop{N}_{0}}{N} \right)=\text{ }\\!\\!\lambda\\!\\!\text{ t} \left\\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\\}
Decay constant  !!λ!! =1tln(N0N)\text{ }\\!\\!\lambda\\!\\!\text{ }=\dfrac{1}{t}\ln \left( \dfrac{\mathop{N}_{0}}{N} \right)
 !!λ!! =15ln(10.998)\Rightarrow \text{ }\\!\\!\lambda\\!\\!\text{ }=\dfrac{1}{5}\ln \left( \dfrac{1}{0.998} \right)

Since we have for half-life there is a relation between decay constant and half life period as given below:
t1/2=0.693λ\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }
Where t1/2{{t}_{{}^{1}/{}_{2}}} is half life period of substance.
On substituting value of decay constant  !!λ!! =4.004005×104years\text{ }\\!\\!\lambda\\!\\!\text{ }=\text{4}\text{.004005}\times \text{1}{{\text{0}}^{-4}}years.
t1/2=0.6934.004005×104years1\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{4.004005\times {{10}^{-4}}year{{s}^{-1}}}
t1/2=1730.767years\Rightarrow {{t}_{{}^{1}/{}_{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=NoeλtN={{N}_{o}}{{e}^{-\lambda t}}
When t=t1/2t={{t}_{{}^{1}/{}_{2}}} , N=N02N=\dfrac{{{N}_{0}}}{2} because for half life period remaining amount of substance is half of initial amount of substance.
N02=N0eλt1/2\Rightarrow \dfrac{{{N}_{0}}}{2}={{N}_{0}}{{e}^{-\lambda {{t}_{{}^{1}/{}_{2}}}}}
12=eλt1/2\Rightarrow \dfrac{1}{2}={{e}^{-\lambda {{t}_{{}^{1}/{}_{2}}}}}
We can write it in terms of natural logarithmic function as
ln(12)=λt1/2\Rightarrow \ln \left( \dfrac{1}{2} \right)=-\lambda {{t}_{{}^{1}/{}_{2}}}
ln(12)=λt1/2\Rightarrow -\ln \left( \dfrac{1}{2} \right)=\lambda {{t}_{{}^{1}/{}_{2}}}
ln(2)=λt1/2\Rightarrow \ln \left( 2 \right)=\lambda {{t}_{{}^{1}/{}_{2}}} \left\\{ \because -\ln \left( \dfrac{a}{b} \right)=\ln \left( \dfrac{b}{a} \right) \right\\}
t1/2=ln(2)λ\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{\ln (2)}{\lambda }
t1/2=0.693λ\Rightarrow {{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\lambda }