Respuesta :
Answer is: the half-life of the radioisotope is 74 days.
m₀ = 3.25 g.
m₁ = 1.21 g.
t = 105 d.
ln(m₀/m₁) = k· t.
ln(3.25/1.21) = k·105 d.
ln(2.685) = 105·k.
0.98 = 105k.
k = 0.0094.
t1/2 = ln2 / k.
t1/2 = 0.693 / 0.0094.
t1/2 = 73.72 days.
m₀ = 3.25 g.
m₁ = 1.21 g.
t = 105 d.
ln(m₀/m₁) = k· t.
ln(3.25/1.21) = k·105 d.
ln(2.685) = 105·k.
0.98 = 105k.
k = 0.0094.
t1/2 = ln2 / k.
t1/2 = 0.693 / 0.0094.
t1/2 = 73.72 days.
The half-life of the radioisotope : 73.68 days
Further explanation
The core reaction is a reaction that causes changes in the core structure.
In the core reaction, the number of atomic numbers and mass numbers of the reactants is the same as the atomic number and mass number of the product
Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles, among others
- alpha α particles ₂He⁴
- beta β ₋₁e⁰ particles
- gamma particles γ
- positron particles ₁e⁰
General formulas used in decay:
[tex] \large {\boxed {Nt=No(\frac {1} {2})^{\frac {T} {t1 / 2}}}} [/tex]
T = duration of decay
t 1/2 = half-life
No = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
A sample of iridium-192 is initially 3.25 g and 1.21 g remains after 105 days, estimate the half-life of the radioisotope
So :
No = 3.25 g
Nt = 1.21 g
T = 105 days
We input this to equation to find half life ([tex]\displaystyle t\frac{1}{2}[/tex]
[tex]\displaystyle Nt=No(\frac {1} {2})^{\frac {T} {t\frac{1}{2}}[/tex]
[tex]\displaystyle 1.21=3.25(\frac {1} {2})^{\frac {105} {t\frac{1}{2}}[/tex]
[tex]\displaystyle 0.372=(\frac{1}{2})^{\frac{105}{t\frac{1}{2} }[/tex]
[tex]\displaystyle log~0.372=\frac{105}{t\frac{1}{2} }.log~0.5\\\\-0.429=\frac{105}{t\frac{1}{2} }.-0.301\\\\1.425=\frac{105}{t\frac{1}{2} }\\\\t\frac{1}{2}=\boxed{\bold{73.68~days}}[/tex]
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Keywords: the half-life, decay reaction, iridium-192
