Find the half-life of U²³⁸, if one gram of it emits [latex]1.24\times 10^4\ \alpha[/latex] – particles per second. (Avogadro’s number = [latex]6.023\times 10^{23}[/latex]).

Solution

Given,

Rate of decay [latex](\frac{dN}{dt})[/latex] = 1.24 x 10⁴ dis/sec

Half life [latex](T_{1/2})[/latex] =?

Mass (m) = 1 gram

Using Avogadro’s hypothesis,

[latex]\because[/latex] 238 gram of U²³⁸ = 6.023 x 10²³ atoms of it

[latex]\therefore[/latex] 1 gram of U²³⁸ = [latex]\frac{6.023\times 10^{23}}{238} = 2.53\times 10^{21}[/latex] atoms

N = 2.53 x 10²¹ atoms

Now, we know that,

[latex]\frac{dN}{dt} = \lambda N[/latex]

Or, 1.24 x 10⁴ = [latex]\lambda[/latex] x (2.53 x 10²¹)

Or, [latex]\lambda = 4.9\times 10^{-18}[/latex] per second