Chemistry: A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there are 100 milligrams of the material present and if after two years it is observed that 5 percent of the original mass has decayed, find (a) an expression for the mass at any time t and (b) the time necessary for 10 percent of the original mass to have decayed.

A certain radioactive material is known to decay

Ответ:

Dragonskeld

21.01.2022

Chemistry

A general equation is given below; it takes two t1/2 to decay 10% of the material.

Explanation:

Decay constant, k = $\frac{ln 2}{t_{1/2} }$

But $ln (\frac{N_{t} }{N_{o} })$ = -kt

Where Nf = final amount

No = initial amount

t = time elapsed

a.)

Initially, 5% of 100mg has decayed:

ln $\frac{95}{100}$ = -k × 2 years

k = -0.051293 ÷ (-2)

=0.0256

⇒ The equation for mass at any time t for this material, Nt :

$N_{t} = 95. e^{-0.0256t}$

b.)

at 10% decay of 100mg,

Nt = 90mg

⇒ $ln (\frac{90}{100})$ = -0.0256 × t

∴ t = 4.108 ≅ 4.1 years (or 2 half lives)

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Ответ:

nataliahenderso

21.12.2022

Chemistry

(See explanation)

Step-by-step explanation:

The differential equation for a exponential growth is:

$\frac{dn}{dt} = k\cdot n$

The formula is rearranged and integrated afterwards:

$\frac{dn}{n} = k \cdot dt$

$\int\limits_{n_{o}}^{n} {\frac{dn}{n} } = k\cdot \int\limits_{0}^{t}\, dt$

$\ln \frac{n}{n_{o}} = k\cdot t$

$\frac{n}{n_{o}} = e^{ k\cdot t}$

$n = n_{o}\cdot e^{k\cdot t}$

The exponential growth function can be used to find the montly growth function as follows:

$\ln 1.2 = k\cdot 12$

k = 0.015

The coeffcients of the expression are, respectively:

$n_{o} = 25$

k = 0.015

The exponential growth function with a monthly growth rate is:

$n(t) = 25\cdot e^{0.015\cdot t}$ , where t is measured in months.

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