Question

Relation between half cycle and decay constant

Answer 1

in a radioactive sample, number of atoms at any instant is given by,

$N=N_0e^{-\lambda t}$

now when t = $T_{1/2}$,

then, $N=N_0/2$

$N_0/2=N_0e^{-\lambda T_{1/2}}$

or, $\frac{1}{2}=e^{-\lambda T_{1/2}}$

or, $-ln2=-\lambda T_{1/2}$

or, $\boxed{\bf{T_{1/2}=\frac{ln2}{\lambda}}}$ This is the required expression which is relation between half cycle and decay constant.

Answer 2

The beginning of the beginning (t = 0) is the number of atoms in any radioactive substance (No) , and In( t) time, the number of remaining united atoms in the substance remains (N)

According to Rutherford and Sodi rules,

$\boxed{\huge{ \bf \: N = N _{0}e {}^{ - \lambda \: t} }}$

if the substance is semi-age t, then t = T after the time the number of remaining united atoms is either,

$\bf \: N \: = \frac{N _{0} }{2}$

$\bf \: or \: \: \frac{N _{0}}{2} = N _{0}e {}^{ - \lambda \: T} \\ \\ \\ \implies \: \: \bf \: \frac{1}{2} = e {}^{ - \lambda \:T } \\ \\ \\ \implies \bf \: e {}^{ \lambda \: T} = 2$

let the relationship between the two logos,

$\bf \lambda \: T \: = log_{e}(2) = 0.6931 \\ \\ \\ or \: \: \bf \: \: \boxed{ T \: = \frac{0.6931}{ \lambda} }$

This is semi-age (T) and decay determinant ($\lambda$).