The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How much of the sample remains after 120 years? (Round your answer to mg (c) After how long will only 1 mg remain? (Round your answer to one decimal place.) T = yr

a)

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Applying ln to both sides of the equality.

So

180-mg sample, so Q(0) = 180

(b) How much of the sample remains after 120 years?

This is Q(120).

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

Applying ln to both sides

225.8 years.

(a) The mass remains after t years is mg.

(b)Therefore the remains sample after 120 years is 11.25 mg.

(c)Therefore after 224.76 years only 1 mg will remain.

Step-by-step explanation:

The differential equation of decay

Integrating both sides

            [ is arbitrary constant ]

Initial condition is, when t=0

Therefore ........(1)

N=  Amount of radioactive material after t unit time.

= initial amount of radioactive material

k= decay constant.

Half life:

, t= 30 years

(a)

The mass remains after t years N.

Now we put the value of in the equation (1)

.........(2)

The mass remains of cesium after t years is mg.

(b)

Putting and t=120 years in equation (2)

Therefore the remains sample after 120 years is 11.25 mg.

(c)

Now putting N= 1 in equation (2)

Taking ln both sides

⇒t=224.76 (approx)

Therefore after 224.76 years only 1 mg will remain.

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)

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step-by-step explanation: