Question

The relation between  \(λ\) and \(T_\frac 12\)is: 
\((T_\frac 12 = half\  life, \ λ→decay\  constant)\)

• A. \(T_{\frac 12}=\frac {ln\ 2}{λ}\)
• B. \(T_{\frac12} ln\ 2=λ\)
• C. \(T_{\frac12}=\frac 1λ\)
• D. \((λ+T_{\frac 12})=\frac {ln\ 2}{2}\)

Answer 1

The Correct Option is A

The relationship between the decay constant \(λ\) and the half-life \(T_{\frac{1}{2}}\) of a radioactive sample is determined by the exponential decay law. The formula linking these two quantities is: 

\(T_{\frac{1}{2}} = \frac{\ln 2}{λ}\)

Here is the step-by-step derivation and explanation:

Half-life is defined as the time required for a quantity to reduce to half its initial amount. For a radioactive substance, this can be expressed using the decay law:

Exponential decay is given by the equation:

\(N(t) = N_0 \cdot e^{-λt}\)

where \(N(t)\) is the quantity remaining after time \(t\), \(N_0\) is the initial quantity, and \(λ\) is the decay constant.

To find \(T_{\frac{1}{2}}\), set \(N(T_{\frac{1}{2}}) = \frac{N_0}{2}\):

\(\frac{N_0}{2} = N_0 \cdot e^{-λT_{\frac{1}{2}}}\)

Simplify by dividing both sides by \(N_0\):

\(\frac{1}{2} = e^{-λT_{\frac{1}{2}}}\)

Take the natural logarithm on both sides to solve for \(T_{\frac{1}{2}}\):

\ln \left(\frac{1}{2}\right) = -\ln 2, as -\ln 2 = -λT_{\frac{1}{2}}

Finally, solve for \(T_{\frac{1}{2}}\) by dividing by T_{\frac{1}{2} = \frac{\ln 2}{λ}

The correct answer to the question is clearly \(T_{\frac{1}{2} = \frac{\ln 2}{λ}\), matching option 1.