Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

Verify the three Q values reported for the reactions given in Fig. 43-11.The needed atomic and particle masses are
$\begin{array}{l} {}^{1}H e & 1.007825 u & {}^{4}H e & 4.002603 u \\ {}^{2}H & 2.014102 u & e^{\pm} & 0.0005486 u \end{array} {}^{3}H e 3.016029 u$
(Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.)

All the energy values are verified.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Describe the expression for energy

The expression for energy is given by,

$Q = - ∆ m c^{2}$

Here, $Q$ is the energy released in a reaction, $∆ m$ is the mass difference between the parent nuclei and the daughter nuclei, and is the velocity of light.

Step 2: Verify the three Q values reported for the reactions

Consider the first reaction.

${}^{1}H + {}^{1}H \to^{2} H + e^{+} + v$

The energy equation can be written as follows.

$Q_{1} = (2 m_{p} - m_{d} - m_{e^{+}}) c^{2}$

Here, $m_{p}$ is the mass of the proton, $m_{d}$ is the mass of deuteron, and $m_{e^{+}}$ is the mass of positron.

Rewrite the equation as follows.

role="math" localid="1661923054546" $Q_{1} = [2 (m_{1 H} - m_{e^{+}}) - (m_{2 H} - m_{e^{+}}) - m_{e^{+}}] c^{2} = [2 m_{1 H} - m_{2 H} - 2 m_{e^{+}}] c^{2} . . . . . . (1)$

Substitute all the known values in equation (1).

$Q_{1 =} [2 (1.007825 u) - 2.014102 u - 2 (0.0005486 u)] (931.5 M e V / u) = 0.41984568 M e V \approx 0.42 M e V$

Therefore, the energy released is $0.42 M e V$ .

Consider the second reaction.

${}^{2}H + {}^{1}H \to^{3} H e + y$

The energy equation can be written as follows.

$Q_{2} = (m_{2 H} + m_{1 H} - m_{3 H e}) c^{2} . . . . . (2)$

Substitute all the known values in equation (2).

$Q_{2} = [(2.014102 u) + (1.007825 u) - (3.016029 u)] (931.5 M e V / u) = 5.493987 M e V \approx 5.49 M e V$

Therefore, the energy released is $5.49 M e V$ .

Consider the third reaction.

${}^{3}H e + {}^{3}H e \to^{4} H e + = +^{1} H +^{1} H$

The energy equation can be written as follows.

$Q_{3} = [2 (m_{3 H e}) - m_{4 H e} - 2 (m_{1 H})] c^{2} . . . . . . (3)$

Substitute all the known values in equation (3).

$Q_{3} = [2 (3.016029 u) - (4.002603 u) - (1.007825 u)] (931.5 M e V / u) = 12.8593575 M e V \approx 12.86 M e V$

Therefore, the energy released is $12.86 M e V$ .

Therefore, all the energy values are verified.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Describe the expression for energy

Step 2: Verify the three Q values reported for the reactions

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Fundamentals Of Physics

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Short Answer

Verify the three Q values reported for the reactions given in Fig. 43-11.The needed atomic and particle masses areH1e1.007825uH4e4.002603uH22.014102ue±0.0005486uH3e 3.016029u(Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.)

Step by Step Solution

Step 1: Describe the expression for energy

Step 2: Verify the three Q values reported for the reactions

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