You visited us 1 times! Enjoying our articles? Unlock Full Access!

Uses of Radioactive Substances

The activity ...

Question

The activity of a sample of radioactive material is $R_{1}$ at time $t_{1}$ and $R_{2}$ at time $t_{2} (t_{2} > t_{1})$ . Its mean life is T. Then

R_{1} t_{1} = R_{2} t_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R_{1} - R_{2}}{t_{2} - t_{1}} = c o n s t a n t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{2} = R_{1} e^{(t_{1} - t_{2}) / T}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R_{2} = R_{1} e^{(t_{1} / t_{2}) / T}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $R_{2} = R_{1} e^{(t_{1} - t_{2}) / T}$
We have present activity $(N)$ given as $N = N_{0} e^{- t / T}$
where $N_{0}$ is initial activity.
Given that $R_{1} = N_{0} e^{- t_{1} / T}$ and $R_{2} = N_{0} e^{- t_{2} / T}$
Dividing both we get $\frac{R_{1}}{R_{2}} = e^{(t_{2} - t_{1}) / T}$
or $R_{2} = R_{1} e^{(t_{1} - t_{2}) / T}$
Concept used is $\frac{a^{b}}{a^{c}} = a^{b - c}$ and $1 / e^{a} = e^{- a}$
Option C is correct.

Suggest Corrections

Similar questions

Q. The radioactivity of a sample is

R_{1}

at time

T_{1}

and

R_{2}

at time

T_{2}

. If the half life of the specimen is

T

, number of atoms that have disintegrated in time

(T_{2} - T_{1})

is proportional to

The radioactivity of a sample is R₁ at a time T₁ and R₂ at a time T₂. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time (T₂-T₁) is proportional to:

The half life of a ratio active source is T. If radio activities of the samples are $R_{1} - R_{2}$ at time

$T_{1} a n d T_{2}$ Respectively, then the number of atoms disintegrated in time $(T_{2} - T_{1})$ is proportional to

Q. The half life of a radioactive sample is

T .

If the activities of the sample at time

t_{1}

and

t_{2}

(t_{1} < t_{2})

are

R_{1}

and

R_{2}

respectively, then the number of atoms disintegrated in time

t_{2} - t_{1}

is proportional to-

Q. The activity of a radioactive substance is

R_{1}

at time

t_{1}

and

R_{2}

at time

t_{2} (> t_{1})

. Its decay constant is

λ

. Then:

Join BYJU'S Learning Program

The activity of a sample of radioactive material is R1 at time t1 and R2 at time t2(t2>t1). Its mean life is T. Then

The correct option is C R2=R1e(t1−t2)/TWe have present activity (N) given as N=N0e−t/Twhere N0 is initial activity.Given that R1=N0e−t1/T and R2=N0e−t2/T Dividing both we get R1R2=e(t2−t1)/Tor R2=R1e(t1−t2)/TConcept used is abac=ab−c and 1/ea=e−aOption C is correct.

The activity of a sample of radioactive material is $R_{1}$ at time $t_{1}$ and $R_{2}$ at time $t_{2} (t_{2} > t_{1})$ . Its mean life is T. Then