N / N0=e^-2. For t=10, N / N0=0.1633. (a) Find i. (b) For what t will N / N0=1 / 2 ?

Name: Problem 74, Chapter 9: Schaum's Outline of Theory and Problems of Basic Mathematics with Applications to Science and Technology
Uploaded: 2023-11-26T21:16:14-08:00
Duration: 6 min 20 s
Channel: Willis James
Description: N / N0=e^-2. For t=10, N / N0=0.1633. (a) Find i. (b) For what t will N / N0=1 / 2 ?

step 1 Hello welcome to this lesson. In this lesson we show that the amount of radioactive material rem

N / N0=e^-2. For t=10, N / N0=0.1633. (a) Find i. (b) For...

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Key Concepts

Decay Constant

The decay constant, often denoted by k or i in different contexts, is a parameter that indicates the rate of exponential decay. It is determined by setting up the decay equation with known values and solving for the constant using logarithms, typically through the formula k = –(ln(N/N?))/t.

Half-Life

The half-life is defined as the time required for a quantity undergoing exponential decay to reduce to half of its initial amount. It can be calculated by setting N/N? = 1/2 in the exponential decay equation and solving for time, yielding t = ln(2)/k.

Exponential Decay

Exponential decay describes processes where a quantity decreases at a rate proportional to its current value. This behavior is modeled by equations of the form N/N? = e^(–kt), where the decay constant k characterizes how quickly the quantity diminishes over time.

Natural Logarithm

The natural logarithm, with base e, is essential in solving exponential equations because it is the inverse of the exponential function. It enables us to isolate the decay constant or time variable in equations like N/N? = e^(–kt), by applying the logarithmic property ln(e^(–kt)) = –kt.

Transcript

00:01 Hello welcome to this lesson.

00:03 In this lesson we show that the amount of radioactive material remaining in an initially pure sample is given by n of t that is equals to n naught times e to the power negative lambda t where n naught is the initial amount of substance.

00:28 We have lambda which is equals to the decay constant and we have t which is equals to time.

00:45 Of course we have n of t which is equals to the amount of substance at time t.

00:57 So amount of substance at time t.

01:03 Alright so let's begin.

01:08 So we begin by looking at the differential equation dn dt which is equals to negative lambda n.

01:21 So this equation states that the rate of change of radioactive material with respect to time is proportional to the amount of radioactive material present at a time.

01:36 So with this we can do separation of variables.

01:38 We can have the n on n which is equals to negative lambda dt.

01:46 And at this point we can take the integral on both sides.

01:51 So we are basically solving the differential equation.

01:56 Let me bring lambda out because lambda is a constant.

02:01 Ok.

02:02 As such we would have ln n which is equals to negative lambda t plus some constant c.

02:12 And at this point we can take the e of both sides.

02:16 E is the base of logarithm...