WBJEE · Chemistry · Chemical Kinetics
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
- A \(0.693 / 2\)
- B \((0.693)^{1 / 2}\)
- C \((0.693)^2\)
- D 0.693
Answer & Solution
Correct Answer
(B) \((0.693)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Hint: For a radio decay \(T_{1 / 2}=\frac{0.693}{\lambda}\) If \(T_{1 / 2}=\lambda=x\) then \(\mathrm{x}=\frac{0.693}{\mathrm{x}}\) \(\Rightarrow \mathrm{x}^2=0.693, \quad \Rightarrow \mathrm{x}=\mathrm{T}_{1 / 2}=\lambda=(0.693)^{1 / 2}\)
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