Question

Starting from 10g of a radioactive elements, 025 g was left after 5 years. Calculate

i) Rate constant for the decay of the radioactive element.

ii) The amount left after one year.

iii) The time required for half of the element to decay.

iv) Average life of the element.

(A.I.S.B. 2011)
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Answer 1

Solution:

$\red\bigstar \;\boxed{k= \dfrac{2.303}{t}{\log}\dfrac{[A]_0}{[A]}}$

i) Rate constant for the decay of the radioactive element.

here,

• [A]₀ = 10g ,
• [A] = 0.25 g ,
• t = 5 years

Putting the values ,

$\implies k= \dfrac{2.303}{5}\log\dfrac{10}{0.25}\\\\\implies k = \dfrac{2.303}{5}\log40 \\\\\implies \underline{\pink{k=0.7379\sf \;years^{-1}}}$

ii) Amount left after 1 year

• t = 1 year
• [A]₀ = 10 g
• [A] = ?

$\red\bigstar \;\boxed{k= \dfrac{2.303}{t}{\log}\dfrac{[A]_0}{[A]}}$

Putting the values,

$\implies 0.7379 = \dfrac{2.303}{1}\log\dfrac{10}{[A]} \\\\\implies \log\dfrac{10}{[A]}=\dfrac{0.7379\times 1}{2.303}\\\\\implies \log\dfrac{10}{[A]}= 0.3204 \\\\\implies \dfrac{10}{[A]} = \rm{ Antilog}\; (0.3204)\\\\\implies \dfrac{10}{[A]}= 2.09 \\\\\implies [A] = \dfrac{10}{2.09} \\\\ \therefore [A] = 4.785 \; g$

iii) The time required for half of the element to decay.

$\red\bigstar\;\boxed{t_{1/2}=\dfrac{0.693}{k}} \\\\\\\implies t_{1/2}= \dfrac{0.693}{0.7379}\\\\\implies t_{1/2} = 0.9392 \;\sf years$

iv) Average life of the element.

$\red\bigstar\; \tau = \dfrac{1}{k}\\\\\implies \tau = \dfrac{1}{0.7379}\\\\\implies \tau = 1.3552 \;\sf years$