{C^{14}} has half life 5700 years. At end of 11400 years, actual amount left is

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13.Nuclei

easy

${C^{14}}$ has half life $5700$ years. At the end of $11400$ years, the actual amount left is

A

$0.5$ of original amount

B

$0.25$ of original amount

C

$0.125$ of original amount

D

$0.0625$ of original amount

Solution

(b) $N = {N_0} \times {\left( {\frac{1}{2}} \right)^{11400/5700}}$

$ = {N_0}{\left( {\frac{1}{2}} \right)^2}$ $ = 0.25{N_0}$.

Standard 12

Physics

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