Question

49 After decaying for 48 hours, l/16 of the original mass of a radioisotope sample remains unchanged. What is the half-life of this radioisotope?
(1) 3.0 h (3) 12 h(2) 9.6 h (4) 24 h

Answer 1

1/16 left would imply 4 half lives have passed (1/2 left = 1, 1/4 left = 2, 1/8 left = 3, 1/16 left = 4). So 4 half lives passed in 48 hours, meaning dividing 48 by 4 will give you the length of 1 half life. Which in this case is 12 hours.

Answer 2

Answer:

The half life of the radioisotope is 12 hours.

Explanation:

Initial mass of the radioisotope = x

Final mass of  the radioisotope = $\frac{1}{16}\times x$ = 0.0625x

Half life of the radioisotope =$t_{\frac{1}{2}}$

Age of the radioisotope = t = 48 hours

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

Now put all the given values in this formula, we get

$0.0625x=x\times e^{-(\frac{0.693}{t_{1/2}})\times 48 h}$

$\ln(0.0625) \times \frac{1}{(-0.693\times 48 h)}=\frac{1}{t_{1/2}}$

$t_{1/2} = 11.99 hours = 12 hours$

The half life of the radioisotope is 12 hours.