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

Rate of Change

The activity ...

Question

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:

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

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

R_{2} = R_{1} e^{λ (t_{2} - t_{1})}

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

R_{2} = R_{1} e^{λ (t_{1} - t_{2})}

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

\frac{R_{1} - R_{2}}{t_{2} - t_{1}} =

constant

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

Open in App

Solution

The correct option is D $R_{2} = R_{1} e^{λ (t_{1} - t_{2})}$
Radioactivity :
$R (t) = R_{o} e^{- λ t}$
$(1) R_{1} = R_{o} e^{- λ t_{1}}$
$(2) R_{2} = R_{o} e^{- λ t_{2}}$
From $(1)$ ,
$R_{o} = R_{1} e^{λ t_{1}}$
Putting in $(2)$ ,
$R_{2} = R_{1} e^{λ (t_{1} - t_{2})}$

Suggest Corrections

Similar questions

R_{1}

t_{1}

R_{2}

t_{2} (t_{2} > t_{1})

\frac{R_{2}}{R_{1}}

R_{1}

t_{1}

R_{2}

t_{2}

t_{2} > t_{1}

λ

t_{1}

t_{2}

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

R_{1}

t_{1}

R_{2}

t_{2} (t_{2} > t_{1})

R_{1}

T_{1}

R_{2}

T_{2}

T

(T_{2} - T_{1})

Join BYJU'S Learning Program