Question

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

Answer 1

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

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

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

(c) Number of remaining sample in 14.6 days is :

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.