Respuesta :

jushmk
[tex] N_{t} = N_{o} e^{ \frac{-693t}{ t_{ \frac{1}{2} } } } [/tex]

Where,
[tex] N_{t} = Quantity remaining after some time[/tex]
[tex] N_{o} = Original qunatity[/tex]
[tex] t_{ \frac{1}{2} } = Half life = 1.251* 10^{9} years for potassium-40[/tex]

Substituting;
[tex] N_{} =800* e^{ \frac{-0.693*3.9* 10^{9} }{1.251* 10^{9} } } [/tex] = 151.83 g

Ans: Amount of K-40 remaining = 92.2 g

Given:

Initial mass of potassium-40 = M₀ = 800 g

Time, t = 3.9 *10⁹ years

To determine:

The mass of potassium-40 (M(t)) that will remain after time 't'

Explanation:

The radioactive decay equation is given as:

[tex]\frac{M(t)}{M_{0} } = e^{\frac{-0.693t}{t_{1/2} } }[/tex] ----(1)

where t1/2 is the half-life of K-40 = 1.251*10⁹ years

Substitute for M₀, t and t1/2 in equation(1):

[tex]M(t) = 800 e^{\frac{-0.693*3.9*10^{9} }{1.251*10^{9} } } = 92.2 grams[/tex]

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