Question

Lt is known that the decay rate of radium is directly proportional to its quantity at each given instant Find the law of variation of a mass of radium as a function of time if at t = 0 , the mass of the radius was $m_0$ and during time $t_0\alpha$% of the original mass of radium decay, if $m=m_0{e}^{-kt}$, then k=?

• A. $k=\frac{1}{t_0}\ln (1-\frac{\alpha }{100})$
• B. $k=\frac{-1}{t_0}\ln (1-\frac{\alpha }{100})$
• C. $k=\frac{1}{t_0}\ln (1-\alpha )$
• D. $k=\frac{-1}{t_0}\ln (1-\alpha )$

Answer 1

The correct option is B $k=\frac{-1}{t_0}\ln (1-\frac{\alpha }{100})$

$\frac{dm}{dt}=-k\left(m_0{e}^{-kt}\right)$
$\frac{dm}{dt}=-km$ Or
$\frac{dm}{dt}\propto m$ ... Law of radioactive decay.
Now $\frac{dm}{m}=-k.dt$
${\int }_{m_0}^m\frac{dm}{m}=-k{\int }_0^{t_0}.dt$
$ln\frac{m}{m_0}=-k{t}_0$ ...(a)
Now
Hence % remaining
$=(1-\frac{\alpha }{100}).m_0$
$=m$
Or $\frac{m}{m_0}=1-\frac{\alpha }{100}$
Substituting in (a) we get
$ln(1-\frac{\alpha }{100})=-k{t}_0$
Or $k=\frac{-1}{t_0}ln(1-\frac{\alpha }{100})$ ...(i)
Now
$k=\frac{ln2}{t_{\frac{1}{2}}}$
Now both $ln\left(2\right)$ and $t_{\frac{1}{2}}$ are positive values. Thus the decay constant K is always positive.
Considering 1,
$0<1-\frac{\alpha }{100}<1$
Hence $ln(1-\frac{\alpha }{100})<0$
This is balanced by the negative sign to make the decay constant positive.
Hence $k=\frac{-1}{t_0}ln(1-\frac{\alpha }{100})$