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


Sagot :

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:

The half life of the radioisotope is 12 hours.

Explanation:

Initial mass of the radioisotope = x

Final mass of  the radioisotope = [tex]\frac{1}{16}\times x[/tex] = 0.0625x

Half life of the radioisotope =[tex]t_{\frac{1}{2}}[/tex]

Age of the radioisotope = t = 48 hours

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope

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

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope

[tex]\lambda[/tex] = rate constant

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

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

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

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

[tex]t_{1/2} = 11.99 hours = 12 hours[/tex]

The half life of the radioisotope is 12 hours.