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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

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Payal Gupta | Contributor-Level 10

3 months ago

13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope = $N_{o}$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of $N_{o}$ remains after decay. Hence, we can write,

$\frac{N}{N_{o}}$ = 3.125% = $\frac{3.125}{100}$ = $\frac{1}{32}$

But $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{32}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{32}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 32$

$- λ t$ = 0 – 3.465

$t$ = $\frac{3.465}{λ}$

Since $λ$ = $\frac{0.693}{T}$

t = $\frac{3.465}{\frac{0.693}{T}}$ = 5T years

Hence, all the isotopes will take about 5T years to reduce 3.125% of its original value.

After decay, the amount of radioactive isotope = N

It is gi

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13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope = $N_{o}$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of $N_{o}$ remains after decay. Hence, we can write,

$\frac{N}{N_{o}}$ = 3.125% = $\frac{3.125}{100}$ = $\frac{1}{32}$

But $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{32}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{32}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 32$

$- λ t$ = 0 – 3.465

$t$ = $\frac{3.465}{λ}$

Since $λ$ = $\frac{0.693}{T}$

t = $\frac{3.465}{\frac{0.693}{T}}$ = 5T years

Hence, all the isotopes will take about 5T years to reduce 3.125% of its original value.

After decay, the amount of radioactive isotope = N

It is given that only 1.0% of $N_{o}$ remains after decay. Hence, we can write,

$\frac{N}{N_{o}}$ = 1% = $\frac{1}{100}$

But $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{100}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{100}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 100$

$- λ t$ = 0 – 4.605

$t$ = $\frac{4.605}{λ}$

Since $λ$ = $\frac{0.693}{T}$

t = $\frac{4.605}{\frac{0.693}{T}}$ = 6.645T years

Hence, all the isotopes will take about 6.645T years to reduce 1.0% of its original value.

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13.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

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