Question

(d) 4
A radioisotope 41 Ar initially decays at
the rate of 34,500 disintegrations per
minute, but decay rate falls to 21,500
disintegrations per minute after 75
minutes. The t'/2 for 41 Ar is
(a) 90 minutes
(b) 110 minutes
180 minutes
(d) 220 minutes​

Answer 1

Explanation:

your answer

A radiistop 41 Ar........

...................b) 110 minutes

option (b )is correct

Answer 2

The $\frac{t'}{2}$ for Ar₄₁ is 110 minutes. (Option - b)

Given:

the initial decay rate - r' = 34500 disintegrations/min

the final decay rate - r = 21,500 disintegrations/min

                        time - t = 75 minutes

To Find: The t'/2 for Ar₄₁

Solution:

Formula used:

$ln\frac{r}{r'} = \frac{- 0.693t}{\frac{t'}{2} }$

• The Formula is derived for initial and final amounts of radioactive substance i.e. $ln \frac{N}{N'} = \frac{-0.693t}{\frac{t'}{2} }$ (N - the final amount of radioactive element, N' - initial amount)    but rates of decay can also be used as both mass and rates of disintegration.

Applying the above formula:-

$ln\frac{21500}{34500} = \frac{- 0.693 * 75}{\frac{t'}{2} }$

$ln 0.623 = \frac{-52}{\frac{t'}{2} }$

$-0.47 = \frac{-0.52}{\frac{t'}{2} }$

$\frac{t'}{2} = \frac{- 52}{- 0.47}$

$\frac{t'}{2} = 110.64$ minutes

Hence, the $\frac{t'}{2}$ for Ar₄₁ = 110 minutes