Bundle: Inquiry into Physics, Loose-Leaf Version, 8th + WebAssign Printed Access Card for Ostdiek/Bord's Inquiry into Physics, 8th Edition, Single-Term
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Chapter 11, Problem 3C
To determine

(a)

The number of alpha particles in decay chain starting from U (238) to Pb (206) and U (235) to Pb (207).

Expert Solution
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Explanation of Solution

Given:

The decay chain of U (238) and U (235) is given as,

U238Pb206U235Pb205.

Calculation:

In the decay chain starting from U (238) to Pb (206).

As we know the mass of alpha particle is 4.

Let there are x number of alpha particles in a decay chain.

2384x=20632=4xx=8

Thus, there are 8 alpha particles.

In the decay chain starting from U (235) to Pb (207).

As we know the mass of alpha particle is 4.

Let there are x number of alpha particles in a decay chain.

2354x=20728=4xx=7

Thus, there are 7 alpha particles.

Conclusion:

In decay chain U (238) to Pb (206) there are 8 alpha particles

In decay chain U (235) to Pb (207) there are 7 alpha particles.

To determine

(b)

The ratio of number of Uranium-238 to lead-206.

Expert Solution
Check Mark

Explanation of Solution

Given:

NumberofUraniuminitially,N0U=NumberofLeadinitially,N0Pbtimeofsample,t=4.5billionyears=4.5×109years.

Formula used:

Number of radioactive nuclei at time t is given by,

NtN0=eλt

Where,

Nt = number of radioactive nuclei at time t

N0 = number of radioactive nuclei initially

t = time of sample

λ = decay constant.

Calculation:

For U (238), number of Uranium-238 at time t is given by,

NtUN0U=eλt...(1)

For Pb (206), number of lead-206 at time t is given by

NtPbN0Pb=eλt...(2)NtUNtPb=eλteλtNtUNtPb=1.

Conclusion:

The ratio of number of Uranium-238 to lead-206 is, NtUNtPb=1.

To determine

(c)

The ratio of number of Uranium-235 to lead-205.

Expert Solution
Check Mark

Explanation of Solution

Given:

NumberofUraniuminitially,N0U=NumberofLeadinitially,N0Pbtimeofsample,t=4.5billionyears=4.5×109years.

Formula used:

Number of radioactive nuclei at time t is given by,

NtN0=eλt

Where,

Nt = number of radioactive nuclei at time t

N0 = number of radioactive nuclei initially

t = time of sample

λ = decay constant.

Calculation:

For U (238), Number of radioactive nuclei at time t is given by,

NtUN0U=eλt

NtPbN0Pb=eλtNtUNtPb=eλteλtNtUNtPb=1.

Conclusion:

The ratio of number of Uranium-235 to lead-207 is NtUNtPb=1.

To determine

(d)

The ratio of number of lead-206 to lead-207.

Expert Solution
Check Mark

Explanation of Solution

Formula used:

Number of radioactive nuclei at time t is given by,

NtN0=eλt

Where,

Nt = number of radioactive nuclei at time t

N0 = number of radioactive nuclei initially

t = time of sample

λ = decay constant.

Calculation:

Lead-206 and lead-207 are the most stable atoms. Generally, these atoms do not decay at all.

Thus, the ratio of number of Pb (206) nuclei and number of Pb (207) will be equal to the number of atoms of Pb (206) nuclei and number of Pb (207) present initially.

Nt206Nt207=N0206N0207.

Conclusion:

The ratio of number of lead-206 to lead-207 is Nt206Nt207=N0206N0207.

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Students have asked these similar questions
Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?
* Coulomb's Law Example Three charges are positioned as seen below. Charge 1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is - 6.0MC. What is the magnitude and the direction of the force on charge 2 due to charges 1 and 3? 93 kq92 F == 2 r13 = 0.090m 91 r12 = 0.12m 92 Coulomb's Constant: k = 8.99x10+9 Nm²/C² ✓
Make sure to draw a Free Body Diagram as well

Chapter 11 Solutions

Bundle: Inquiry into Physics, Loose-Leaf Version, 8th + WebAssign Printed Access Card for Ostdiek/Bord's Inquiry into Physics, 8th Edition, Single-Term

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