Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 20P

(a)

To determine

The differential equation that defines the number of daughter nucleus.

(a)

Expert Solution
Check Mark

Answer to Problem 20P

The daughter nucleus satisfies the differential equation is dN2dt=λ1N1λ2N2.

Explanation of Solution

Let N10 be the number of parent nuclei at time t=0, N1(t) be the number of parent nuclei at any instant t and λ1 is the decay constant of the parent nuclei. The number of daughter nucleus at time t=0 is zero, N2(t) be the number of daughter nuclei at any instant t and λ2 is the decay constant of the daughter nuclei.

Write the expression for decay rate

  dNdt=λN                                                                                                  (I)

Here, dNdt is the rate of decay, λ is the decay constant and N is the number of nuclei.

Write the expression for rate of change of the daughter nuclei

  dN2dt=dN2+dtdN2dt                                                                                                  (II)

Here, dN2dt is the rate of change of the daughter nuclei, dN2+dt is the rate of production of the daughter nuclei and dN2dt is the rate of decay of the daughter nuclei.

The rate of decay of the parent nucleus is same as the rate of production of the daughter nuclei.

Substitute dN1dt for dN2+dt in (II)

  dN2dt=dN1dtdN2dt                                                                                                 (III)

Conclusion:

Substitute λ1N1 for dN1dt  and λ2N2 for dN2dt in (III) to find dN2dt.

  dN2dt=λ1N1λ2N2                                                                                                (IV)

Thus, the daughter nucleus satisfies the differential equation is dN2dt=λ1N1λ2N2.

(b)

To determine

The solution to the above differential equation using the verification by substitution method.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

N2(t)=N10λ1λ1λ2(eλ2teλ1t) is a solution to the differential equation.

Explanation of Solution

Write the expression for trail solution

  N2(t)=N10λ1λ1λ2(eλ2teλ1t)                                                                                   (V)

Differentiate the above equation with respect to t

  dN2dt=N10λ1λ1λ2(λ2eλ2t+λ1eλ1t)                                                                            (VI)

Multiply equation (V) by λ2 and add it with (V)

  dN2dt+λ2N2=N10λ1λ1λ2(λ2eλ2t+λ1eλ1t+λ2eλ2tλ2eλ1t)                                      (VII)

Write the expression for N1

  N1=N10eλ1t                                                                                                 (VIII)

Conclusion:

Solve (VII) further

  dN2dt+λ2N2=N10λ1λ1λ2(λ2eλ2t+λ1eλ1t+λ2eλ2tλ2eλ1t)=N10λ1λ1λ2(λ1λ2)eλ1t=N10λ1eλ1t

Substitute (VIII) in the above equation and rearrange.

  dN2dt=λ1N1λ2N2

Thus, N2(t)=N10λ1λ1λ2(eλ2teλ1t) is a solution to the differential equation.

(c)

To determine

The number of P218o and P214b nuclei versus time graph.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

The number of P218o and P214b nuclei versus time graph is given below:

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 30, Problem 20P , additional homework tip  1

Explanation of Solution

Write the expression for decay constant

  λ=ln2T1/2                                                                                                             (IX)

Here, T1/2 is the half-life period.

Substitute 3.10min for T1/2 to find λ1. Here, λ1 is the decay constant of P218o

  λ1=ln23.10 min=0.2236min1

Substitute 26.8min for T1/2 to find λ2. Here, λ2 is the decay constant of P214b

  λ2=ln226.8 min=0.0259min1

Substitute 0.2236min1 for λ1 and 1000 for N10 in (VIII) to find N1

  N1=(1000)e(0.2236min1)t                                                                                        (X)

Substitute 0.0259min1 for λ2, 0.2236min1 for λ1 and 1000 for N10 in (V) to find N2

  N2=(1000)(0.2236min1)(0.2236min10.0259min1)(e(0.0259min1)te(0.2236min1)t)=1131(e(0.0259min1)te(0.2236min1)t)                   (XI)

Conclusion:

Using expression (X) and (XI), table the number of P218o and P214b nuclei for various t values

t (min)N1N2
010000
2445350
4408557
6261673
8167730
10107752
1268.3751
1443.7737
1627.9715
1817.9689
2011.4660
227.30631
244.67602
262.99573
281.91545
301.22519
320.781493
340.499468
360.319445

Using the above data in the table construct the graph

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 30, Problem 20P , additional homework tip  2

Thus, the above graph shows the number of nuclei as a function of time.

(d)

To determine

The instant when the number of P214b is maximum inferred from the above graph.

(d)

Expert Solution
Check Mark

Answer to Problem 20P

The number of P214b is maximum in the time interval t=10min to t=12min.

Explanation of Solution

From the above graph, the P214b is maximum in the time interval t=10min to t=12min.

(e)

To determine

An expression for maximum number of P214b in terms of λ1 and λ2.

(e)

Expert Solution
Check Mark

Answer to Problem 20P

The time for maximum P214b is given by the expression, tm=ln(λ1/λ2)λ1λ2.

Explanation of Solution

Equate (VI) to zero to find the maximum

  dN2dt=N10λ1λ1λ2(λ2eλ2t+λ1eλ1t)=0                                                                 (XII)

Conclusion:

Simplify and rearrange for t

  λ1eλ1t=λ2eλ2tt=tm=ln(λ1/λ2)λ1λ2                                                                                       (XIII)

Here, tm is the time when the number of P214b is maximum.

Thus, time for maximum P214b is given by the expression, tm=ln(λ1/λ2)λ1λ2.

(f)

To determine

The instant when the number of P214b is maximum calculated using the above expression.

(f)

Expert Solution
Check Mark

Answer to Problem 20P

The time when the number of P214b is maximum is 10.9min.

Explanation of Solution

The time for maximum P214b is given in the above part.

Conclusion:

Substitute 0.0259min1 for λ2 and 0.2236min1 for λ1 in (XIII) to find tm

  tm=ln(λ1/λ2)λ1λ2=ln(0.2236min1/0.0259min1)0.2236min10.0259min1=10.9min

Thus, the time when the number of P214b is maximum is 10.9min and it agrees with the answer in part (c).

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Chapter 30 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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