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If in case of a radio isotope the value of half-life \(\left(T_{1/2}\right)\) and decay constant \((\lambda)\) are identical in magnitude, then their value should be
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The correct answer is:
\((0.693)^{1 / 2}\)
Hint: For a radio decay \(T_{1 / 2}=\frac{0.693}{\lambda}\)
If \(T_{1 / 2}=\lambda=x\) then \(\mathrm{x}=\frac{0.693}{\mathrm{x}}\)
\(\Rightarrow \mathrm{x}^2=0.693, \quad \Rightarrow \mathrm{x}=\mathrm{T}_{1 / 2}=\lambda=(0.693)^{1 / 2}\)
If \(T_{1 / 2}=\lambda=x\) then \(\mathrm{x}=\frac{0.693}{\mathrm{x}}\)
\(\Rightarrow \mathrm{x}^2=0.693, \quad \Rightarrow \mathrm{x}=\mathrm{T}_{1 / 2}=\lambda=(0.693)^{1 / 2}\)
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