The decay constant of a radio active substanc | vedclass

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Question

Given $\lambda=0.173$

$T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{0.173} \cong 4$

Also for t $1 / 0.173$ year

Remaining nuclei $N=N_{0} e^{

Vedclass · Accepted Answer

$(A)$ and $(C)$ both

Vedclass · Answer

Given $\lambda=0.173$

$T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{0.173} \cong 4$

Also for t $1 / 0.173$ year

Remaining nuclei $N=N_{0} e^{-1}=0.37 N_{0}$

Decay nuclie $=N_{0}-N=0.63 N_{0}$

The decay constant of a radio active substance is $0.173\, (years)^{-1}.$ Therefore :

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