EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Question
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Chapter 40, Problem 42P

(a)

To determine

The Q value for the given reaction.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The reaction:

  H2+H2H3+H1+Q

Formula used:

Write the reaction.

  H2+H2H3+H1+Q

Write the expression for the Q value, in MeV, for the decay.

  Q=(mimf)c2(931 .5MeVc -2u) .......... (1)

Here, mi is the mass of reactants and mf is the mass of products.

Write the expression for the mass of reactants.

  mi=mH2+mH2 .......... (2)

Here, mH2 is the mass of H2 .

Write the expression for the mass of products.

  mf=mH3+m1H .......... (3)

Here, mH3 is the mass of H3 and m1H is the mass of H1 .

Calculation:

Substitute 2.014102u  for mH2 equation (2).

  mi=2.014102u +2.014102u =4.028204u

Substitute 3.016049u for mH3 and 1.007825u for mn in equation (3).

  mf=3.016049u+1.007825u=4.028204u

Substitute 4.028204u for mi and 4.028204u for mf in equation (1).

  Q=(4.028204u4.028204u)c2( 931 .5MeVc -2 u)=(0.004330u)c2( 931 .5MeVc -2 u)=4.03MeV

Conclusion:

Thus, the Q value for the given reaction is 4.03MeV .

(b)

To determine

The Q value for the given reaction.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The reaction:

  H2+H3eH4e+H1+Q

Formula used:

Write the reaction.

  H2+H3eH4e+H1+Q

Write the expression for the Q value, in MeV, for the decay.

  Q=(mimf)c2(931 .5MeVc -2u) .......... (1)

Here, mi is the mass of reactants and mf is the mass of products.

Write the expression for the mass of reactants.

  mi=mH2+mH3e .......... (2)

Here, mH2 is the mass of H2 and mH3e is the mass of H3e .

Write the expression for the mass of products.

  mf=mH4e+m1H .......... (3)

Here, mH4e is the mass of H4e and m1H is the mass of 1H .

Calculation:

Substitute 2.014102u  for mH2 and 3.016029u for mH3e in equation (2).

  mi=2.014102u +3.016029=5.030131u

Substitute 4.002602u for mH4e and 1.007825u for m1H in equation (3).

  mf=4.002602u+1.007825u=5.010427u

Substitute 5.030131u for mi and 5.010427u for mf in equation (1).

  Q=(5.030131u5.010427u)c2( 931 .5MeVc -2 u)=(0.019703u)c2( 931 .5MeVc -2 u)=18.4MeV

Conclusion:

Thus, the Q value for the given reaction is 18.4MeV .

(c)

To determine

The Q value for the given reaction.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given:

The reaction:

  L6i+nH3+H4e+Q

Formula used:

Write the reaction.

  L6i+nH3+H4e+Q

Write the expression for the Q value, in MeV, for the decay.

  Q=(mimf)c2(931 .5MeVc -2u) .......... (1)

Here, mi is the mass of reactants and mf is the mass of products.

Write the expression for the mass of reactants.

  mi=mL6i+mn .......... (2)

Here, mL6i is the mass of L6i and mn is the mass of n .

Write the expression for the mass of products.

  mf=mH3+mH4e .......... (3)

Here, mH3 is the mass of H3 and mH4e is the mass of H4e .

Calculation:

Substitute 6.015122 for mL6i and 1.008665u for mn in equation (2).

  mi=6.015122u +1.008665u=7.023787u

Substitute 3.016049u for mH3 and 4.002602u for mH4e in equation (3).

  mf=3.016049u+4.002602u=7.018651u

Substitute 7.023787u for mi and 7.018651u for mf in equation (1).

  Q=(7.023787u7.018651u)c2( 931 .5MeVc -2 u)=(0.003510u)c2( 931 .5MeVc -2 u)=3.27MeV

Conclusion:

Thus, the Q value for the given reaction is 3.27MeV .

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