$_{92} U^{238}$ decays to $_{90} T h^{234}$ with half-life $4.5 \times 1 0^{9}$ year. The resulting $_{90} T h^{234}$ is in excited state and hence, emits a further a gamma ray to come to the ground state, with half-life $1 0^{- 8} s$ . A sample of $_{92} U^{238}$ emits $20$ gamma rays per second. In what time, the emission rate will drop to $5$ gamma ray per second ?

Question

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Answer 1

$\frac{d N}{d t} = - λ N$ [ $\frac{d N}{d t}$ rate of decay $N$ = no. of nucleus]

$\Rightarrow \frac{d N}{N} = - λ d t$

$\Rightarrow lo g N / N_{0} = - λ t$ ............(1)

Now,
$λ N_{0} = 20$ and $λ N = 5$

$N / N_{0} = 1 / 4$

Now from (1) we get

$\Rightarrow lo g 1 / 4 = - \frac{lo g 2}{t _{1 / 2}} t$

$\Rightarrow t = - t_{1 / 2} \frac{lo g 2 ^{- 2}}{lo g 2}$

$\Rightarrow t = 2 t_{1 / 2} = 2 \times 4.5 \times 1 0^{9}$

$= 9 \times 1 0^{9}$ Years.

Answer 2

\frac{d N}{d t} = - λ N

[

\frac{d N}{d t}

rate of decay

N

= no. of nucleus]

\Rightarrow \frac{d N}{N} = - λ d t

\Rightarrow lo g N / N_{0} = - λ t

............(1)

Now,

λ N_{0} = 20

and

λ N = 5

N / N_{0} = 1 / 4

Now from (1) we get

\Rightarrow lo g 1 / 4 = - \frac{lo g 2}{t _{1 / 2}} t

\Rightarrow t = - t_{1 / 2} \frac{lo g 2 ^{- 2}}{lo g 2}

\Rightarrow t = 2 t_{1 / 2} = 2 \times 4.5 \times 1 0^{9}

= 9 \times 1 0^{9}

Years.