Question

Half life of $C^{19}$ is 5700 years. find its decay constant?

Answer 1

GIVEN :–

• Half life $\bf \:( t_{ \frac{1}{2} })$ of $\bf C^{19}$ is 5700 years.

TO FIND :–

• Decay constant $(\lambda)$= ?

SOLUTION :–

• We know that –

$\\ \large \implies { \boxed{ \bf \: t_{ \frac{1}{2} } = \dfrac{0.693}{ \lambda}}}$

• Now put the values –

$\\ \large \implies \bf \: 5700= \dfrac{0.693}{ \lambda}$

$\\ \large \implies \bf \: \lambda= \dfrac{0.693}{5700}$

• We should write this as –

$\\ \large \implies \bf \: \lambda= \dfrac{693}{5700000}$

$\\ \large \implies \bf \: \lambda= \dfrac{693}{57 \times {10}^{5} }$

$\\ \large \implies \bf \: \lambda= \dfrac{693}{57} \times {10}^{ - 5}$

$\\ \large \implies \large{ \boxed{ \bf \lambda=12.26\times {10}^{ - 5} \: year ^{ - 1} }}$

• Hence , Option (1) is correct.

Answer 2

Answer:

option (1)

Explanation:

$\boxed{\bf Given,} \\\\half-life \ of \ C^{19} =5700 \ years \\\\ \implies t_{\frac{1}{2}} =5700 \ years \\\\ \boxed{\bf To \ find,} \\\\ decay \ constant, \lambda=? \\\\ \boxed{\bf solution:} \\\\ we \ know, \\\\ \boxed {t_{\frac{1}{2}} = \frac{0.693}{\lambda}}\\\\\\ \implies \lambda=\frac{0.693}{t_{\frac{1}{2}}} \\\\ \lambda=\frac{0.693}{5700} \\\\ \lambda=\frac{693}{5700\times1000} \\\\ \lambda=\frac{693}{57} \times 10^{-5} \\\\ \lambda=12.16 \times 10^{-5} \ year^{-1}$

$\boxed{\bf \therefore the \ decay \ constant \ = \ 12.16\times10^{-5} \ year^{-1} }$