Question

The half life of ³⁴Cl is 1.53sec. How long does it take for 99.9% of sample of ³⁴Cl to decay?
a) 15.25 sec
b) 10.50sec
c) 25.25sec
d) 2.06sec​

Answer 1

Answer:

It takes 15.25 sec to decay 99.9% of the sample of ³⁴Cl.

Explanation:

Given:

t1/2 = 1.53 sec

N₀ = 100

N = 100-99.9 = 0.1

The half-life of a reaction is the time in which the concentration of the reaction reduces to half of its initial value i.e the time taken for half of the reaction to get completed.

The expression for the half-life is given below.

$t1/2 = \frac{0.693}{k}$

where k is the rate of the reaction.

$1.53 = \frac{0.693}{k}$

Therefore,  $k = \frac{0.693}{1.53}$

The equation for the rate constant is given by

$k = \frac{2.303}{t} log\frac{N0}{N}$

Substituting the values,

$\frac{0.693}{1.53} = \frac{2.303}{t} log\frac{100}{0.1}$

$t = \frac{2.303}{0.453} log(1000)$

Therefore, t= 15.25 sec

Answer 2

Given:

half-life of ³⁴Cl - t' = 1.53s

amount of ³⁴Cl = 99.9%

To Find:

time required for 99.9% of the substance to decay - t

Solution:

³⁴Cl left = 100 - 99.9 = 0.1%

The final amount - N = 0.1%

initial amount of ³⁴Cl - N' = 100

Formula used:

t = $\frac{2.303}{k} log \frac{N'}{N}$

t' = $\frac{2.303}{k} log2$ (as N = 100 and N' = 50,hence 100/50 = 1/2)

t' = $\frac{0.693}{k}$

Applying the above formula of half-life (t') to calculate "k":-

k = 0.693/t'

k = 0.693/1.53

k =0.45

Applying the above formula to calculate the time - t taken for 99.9% decay:-

t = $\frac{2.303}{k} log \frac{N'}{N}$

t = $\frac{2.303}{0.45} log\frac{100}{0.1}$

t = 5.12 log1000

t = 5.12 × 3

t = 15.36s

Hence, the time taken for 99.9% of decay of a sample of ³⁴Cl = 15.36s