Question

A radioactive isotope has a half life of T years
how long will it take the activity to reduce to
a) 3.125%
6) 1% of its original value.

Answer 1

Answer:

(a) 3.125%

HOPE THIS WILL HELP YOU

Answer 2

Answer:-

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(a) Now,

$\implies \sf \: \dfrac{N}{ \: N_0} = \bigg( \dfrac{1}{2} \bigg)^{ \frac{t}{T } } \: \: \: ...(1)$

$\implies \sf \: \dfrac{N}{ \: N_0} = 3.125\% = \dfrac{3.125}{100} = \dfrac{1}{32} = \bigg( \dfrac{1}{2} \bigg)^{5} \: \: \: ...(2)$

From the equations (1) and (2) we have,

$\sf \implies \dfrac{t}{T \: } = 5 \implies \bf \: t = 5T$

Therefore, It will take 5 years.

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(b) Here,

$\implies \sf \: \dfrac{N}{ \: N_0} = 1\% = \dfrac{1}{100} \: \: \: ...(3)$

From the equations (1) and (3)

$\sf \implies \bigg( \dfrac{1}{2} \bigg)^{ \frac{t}{T } } = \dfrac{1}{100} \: or \: \: (2)^{ \frac{t}{T}} = 100$

$\implies \sf \: \dfrac{t}{T} log \: 2 = log \: 100$

$\implies \sf \: \dfrac{t}{T} = \dfrac{log \: 100}{log \: 2} = \dfrac{2}{0.3010} = log \: 2 = 6.645$

$\implies \bf \: t = 6.645T$

Therefore it will take 6.645 years .

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