The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ min. The time, (in minute) at which the activity reduces to half its value, is :

Question

Verified by Toppr

Answer 1

Using first order equation   $N = N_{o} e^{- λ t}$
Given : $N = \frac{N _{o}}{e}$ @ $t = 5 m i n$
$∴$    $\frac{N _{o}}{e} = N_{o} e^{- λ (5)}$
Or $- lo g_{e} e = - 5 λ$
$⟹$ $λ = \frac{1}{5}$            $(∵ lo g_{e} e = 1)$
Time at which the activity reduces to half of its initial value is known as half life of the radioactive substance.
Half life $T_{1 / 2} = \frac{lo g _{e} 2}{λ} = 5 lo g_{e} 2$

Answer 2

Using first order equation

N = N_{o} e^{- λ t}

Given :

N = \frac{N _{o}}{e}

@

t = 5 m i n

∴

\frac{N _{o}}{e} = N_{o} e^{- λ (5)}

Or

- lo g_{e} e = - 5 λ

⟹

λ = \frac{1}{5}

(∵ lo g_{e} e = 1)

Time at which the activity reduces to half of its initial value is known as half life of the radioactive substance.
Half life

T_{1 / 2} = \frac{lo g _{e} 2}{λ} = 5 lo g_{e} 2

The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ min. The time, (in minute) at which the activity reduces to half its value, is :

$lo g_{e} \frac{2}{5}$

$\frac{5}{lo g _{e} 2}$

$5 lo g_{10} 2$

$5 lo g_{e} 2$

A freshly made radioactive sample gives a count rate of $8000$ counts per minute. After twenty days, it gives a count-rate of $500$ counts per minute. What is the half-life of the material?

A count rate meter shows a count of $240$ per minute from a given radioactive source. One hour later the meter shows a count rate of $30$ per minute. The half life of the source is:

A piece of bone from an archaeological site is found to give a count rate of $15$ counts per minute. A similar sample of fresh bone gives a count rate of $19$ counts per minute. Calculate the age of the specimen. Given $T_{1 / 2} = 5570 y e a r s$

The activity of a radioactive sample is measured as $N_{0}$ count per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5 m i n$ . The time (in minutes) at which the activity reduces to half its value is

A count rate meter shows a count of $240$ per minute from a given radioactive source. One hour later the meter shows a count rate of $30$ per minute. The half-life of the source is?

Revise with Concepts

Discovery of Radiation

Introduction to Radioactivity

Alpha, Beta and Gamma Emission

Differentiate Between Alpha, Beta and Gamma Radiations

Background Radiations

The activity of a radioactive sample is measured as N0​ counts per minute at t=0 and N0​/e counts per minute at t=5 min. The time, (in minute) at which the activity reduces to half its value, is :

loge​52​

loge​25​

5log10​2

5loge​2

A freshly made radioactive sample gives a count rate of 8000 counts per minute. After twenty days, it gives a count-rate of 500 counts per minute. What is the half-life of the material?

A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half life of the source is:

A piece of bone from an archaeological site is found to give a count rate of 15 counts per minute. A similar sample of fresh bone gives a count rate of 19 counts per minute. Calculate the age of the specimen. Given T1/2​=5570 years

The activity of a radioactive sample is measured as N0​ count per minute at t=0 and N0​/e counts per minute at t=5min. The time (in minutes) at which the activity reduces to half its value is

A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is?

Revise with Concepts

Discovery of Radiation

Introduction to Radioactivity

Alpha, Beta and Gamma Emission

Differentiate Between Alpha, Beta and Gamma Radiations

Background Radiations

The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ min. The time, (in minute) at which the activity reduces to half its value, is :

$lo g_{e} \frac{2}{5}$

$\frac{5}{lo g _{e} 2}$

$5 lo g_{10} 2$

$5 lo g_{e} 2$

A freshly made radioactive sample gives a count rate of $8000$ counts per minute. After twenty days, it gives a count-rate of $500$ counts per minute. What is the half-life of the material?

A count rate meter shows a count of $240$ per minute from a given radioactive source. One hour later the meter shows a count rate of $30$ per minute. The half life of the source is:

A piece of bone from an archaeological site is found to give a count rate of $15$ counts per minute. A similar sample of fresh bone gives a count rate of $19$ counts per minute. Calculate the age of the specimen. Given $T_{1 / 2} = 5570 y e a r s$

The activity of a radioactive sample is measured as $N_{0}$ count per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5 m i n$ . The time (in minutes) at which the activity reduces to half its value is

A count rate meter shows a count of $240$ per minute from a given radioactive source. One hour later the meter shows a count rate of $30$ per minute. The half-life of the source is?