Question

The mean lives of radioactive substances are $1620y$ and $405y$ for $\alpha -$ emission and $\beta -$ emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by $\alpha -$ emission and $\beta -$ emission simultaneously.

• A. $449y$
• B. $399y$
• C. $549y$
• D. $579y$

Answer 1

Answer: (A)

Let at some instant of time $t$ , number of atoms of the radioactive substance are $N$ . It may decay either by α - emission on by β- emission. So, we can write,
$\left(\frac{- d N}{d t}\right)_{n e t}=\left(\frac{- d N}{d t}\right)_{\alpha }+\left(\frac{- d N}{d t}\right)_{\beta }$
If the effective decay constant is $\lambda$ , then
$\lambda N = \lambda _{\alpha } N ⁡ + \lambda _{\beta } N ⁡$
∴ $\lambda = \lambda _{\alpha } + \lambda _{\beta } = \frac{1}{1 6 2 0} + \frac{1}{4 0 5}$
$=\frac{1}{3 2 4}y^{- 1}$
Now, $\frac{ N _{0}}{4} = N ⁡_{0} e ⁡^{- \lambda t}$
∴ $- \lambda t = \text{ln} \left(\frac{1}{4}\right) = - 1 \text{.} 3 8 6$
or $\left(\frac{1}{3 2 4}\right) t = 1 \text{.} 3 8 6$
∴ $t=449y$