Ncert Exercises & Solutions

13.25 A source contains two phosphorous radio nuclides ${ }_{15}^{32} \mathrm{P}$ (T_1/2 = 14.3 days) and ${ }_{15}^{33} \mathrm{P}$ (T_1/2 = 25.3 days). Initially, 10% of the decays come from ${ }_{15}^{33} \mathrm{P}$. How long one must wait until 90% do so?

Hint: $N=N_oe^{-\lambda t}$
Step 1: Find the initial and final number of nuclie in both elements.

$H a l f - l i f e o f {}_{15}^{32}P, T_{1 / 2 =} 14.3 d a y s$
$H a l f - l i f e o f {}_{15}^{33}P, T'_{1 / 2 =} 25.3 d a y s a n d {}_{15}^{33}P n u c l e u s d e c a y i s 10 % o f t h e t o t a l a m o u n t o f d e c a y .$
$T h e s o u r c e h a s i n i t i a l l y 10 % o f {}_{15}^{33}P n u c l e u s a n d 90 % o f {}_{15}^{32}P n u c l e u s .$
$S u p p o s e a f t e r t d a y s, t h e s o u r c e h a s 10 % o f {}_{15}^{32}P n u c l e u s a n d 90 % o f {}_{15}^{33}P n u c l e u s .$
$I n i t i a l l y :$
$N u m b e r o f {}_{15}^{33}P n u c l e u s = N$
$N u m b e r o f {}_{15}^{32}P n u c l e u s = 9 N$
$F i n a l l y :$
$N u m b e r o f {}_{15}^{33}P n u c l e u s = 9 N'$
$N u m b e r o f {}_{15}^{32}P n u c l e u s = N'$

Step 2: Write the equations for both elements.
For ${ }_{15}^{33} \mathrm{P}$ nucleus, we can write the number ratio as:

$\frac{N'}{9 N} = {(\frac{1}{2})}^{\frac{1}{T_{1 / 2}}}$
$N' = 9N {(2)}^{\frac{- 1}{25.3}} . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$
For${ }_{15}^{32} \mathrm{P}$ nucleus,wecanwritethenumberratioas:
$\begin{aligned} \frac{9 N^{\prime}}{N} &=\left(\frac{1}{2}\right)^{\frac{1}{T_{1 / 2}}} \\ 9 N^{\prime} &=N(2)^{\frac{-1}{14. 3}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(2) \end{aligned}$

$Step 3: Find the required time.
On dividing the equation (1) by the equation (2), we get:$ $\frac{1}{9}=9 \times 2\left(\frac{-t}{25.3} +\frac{t}{14.3}\right)$ $\frac{}{}^{(\frac{}{} \frac{}{})}$
$\frac{1}{81} = 2^{(\frac{11 t}{25.3 \times 14.3})}$
$\log 1 - \log 81 = \frac{- 11 t}{25.3 \times 14.3} \log 2$
$\frac{- 11 t}{25.3 \times 14.3} = \frac{0 - 1.908}{0.301}$
$t = \frac{25.3 \times 14.3 \times 1.908}{11 \times 0.301} \approx 208.5 d a y s$
$H e n c e, i t w i l l t a k e a b o u t 208.5 d a y s f o r 90 % d e c a y o f {}_{15}^{33}P .$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions