For a radioactive material, its activity $A$ and rate of change of its activity $R$ are defined as $A=-\frac{d N}{d t}$ and $R=-\frac{d A}{d t}$, where $N(t)$ is the number of nuclei at time $t$. Two radioactive sources $P$ (mean life $\tau$) and $Q$ (mean life $2 \tau$) have the same activity at $t=0$. Their rates of change of activities at $t=2 \tau$ are $R_{P}$ and $R_{Q}$, respectively. If $\frac{R_{P}}{R_{Q}}=\frac{n}{e}$, then the value of $n$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Modern Physics

For a radioactive material, its activity

A

and rate of change of its activity

R

are defined as

A = - \frac{d N}{d t}

and

R = - \frac{d A}{d t}

, where

N (t)

is the number of nuclei at time

t

. Two radioactive sources

P

(mean life

τ

) and

Q

(mean life

2 τ

) have the same activity at

t = 0

. Their rates of change of activities at

t = 2 τ

are

R_{P}

and

R_{Q}

, respectively. If

\frac{R _{P}}{R _{Q}} = \frac{n}{e}

, then the value of

n

Answer: 2

For source P with mean life

τ

A_{P} = A_{0} e^{- t / τ}

R_{P} = \frac{A _{0}}{τ} e^{- t / τ}

For source Q with mean life

2 τ

A_{Q} = A_{0} e^{- t / (2 τ)}

R_{Q} = \frac{A _{0}}{2 τ} e^{- t / (2 τ)}

t = 2 τ

\frac{R _{P}}{R _{Q}} = \frac{\frac{A _{0}}{τ} e ^{- 2}}{\frac{A _{0}}{2 τ} e ^{- 1}} = \frac{2 e ^{- 2}}{e ^{- 1}} = \frac{2}{e}

Therefore,

n = 2

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer