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

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13.Nuclei

hard

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

A

Nearly $63\%$ of the radioactive substance will decay in $(1/0.173)\, year.$

B

half life of the radio active substance is $(1/0.173) \,year.$

C

one -forth of the radioactive substance will be left after nearly $8$ years.

D

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

Solution

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 $=N_{0}-N=0.63 N_{0}$

Standard 12

Physics

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