EBK STUDENT SOLUTIONS MANUAL WITH STUDY
EBK STUDENT SOLUTIONS MANUAL WITH STUDY
10th Edition
ISBN: 9781337520386
Author: Vuille
Publisher: YUZU
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Question
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Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

The mass of the particles on the left side is 8.023829 u.

Explanation of Solution

The mass of the particles on the left side is,

mi=m(H11)+m(L37i)

  • m(H11) is the mass of H11 .
  • m(L37i) is the mass of L37i .

Substitute 1.007825 u for m(H11) and 7.016004 u for m(L37i) .

mi=(1.007825 u)+(7.016004 u)=8.023829 u

Conclusion:

The mass of the particles on the left side is 8.023829 u.

(b)

To determine

The mass of the particles on the right side.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

The mass of the particles on the right side is 8.025594 u.

Explanation of Solution

The mass of the particles on the left side is,

mf=m(B17e)+m(n01)

  • m(B17e) is the mass of B17e .
  • m(n01) is the mass of n01 .

Substitute 7.016929 u for m(B17e) and 1.008665 u for m(n01) .

mf=(7.016929 u)+(1.008665u)=8.025594 u

Conclusion:

The mass of the particles on the right side is 8.025594 u.

(c)

To determine

The Q-value.

(c)

Expert Solution
Check Mark

Answer to Problem 38P

The Q-value is 1.64MeV .

Explanation of Solution

Formula to calculate the Q-value is,

Q=(mimf)c2

  • c is the speed of light.

Substitute 8.023829 u for mi , 8.025594 u for mf and 931.5 MeV/u for c2 .

Q=[(8.023829 u)(8.025594 u)](931.5MeV/u)=1.64MeV

Conclusion:

The Q-value is 1.64MeV .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution
Check Mark

Answer to Problem 38P

The expression describing the law of conservation of momentum is mpv=(mBe+mn)V

Explanation of Solution

Formula to calculate the initial momentum is,

pi=mpv       (I)

Formula to calculate the final momentum is,

pf=mBeV+mnV=(mBe+mn)V       (II)

From Law of conservation of momentum,

pi=pf       (III)

Substitute Equations (I) and (II) in (III).

mpv=(mBe+mn)V

Conclusion:

The expression describing the law of conservation of momentum is mpv=(mBe+mn)V

(e)

To determine

The expression relating the kinetic energies of the particles.

(e)

Expert Solution
Check Mark

Answer to Problem 38P

The expression relating the kinetic energies of the particles is 12mpv2+Q=12(mBe+mn)V2

Explanation of Solution

Formula to calculate the initial kinetic energy is,

KEi=12mpv2+Q       (IV)

Formula to calculate the final kinetic energy is,

KEf=12(mBe+mn)V2       (V)

From Law of conservation of energy,

KEi=KEf       (VI)

Substitute Equations (IV) and (V) in (VI).

12mpv2+Q=12(mBe+mn)V2

Conclusion:

The expression relating the kinetic energies of the particles is 12mpv2+Q=12(mBe+mn)V2

(f)

To determine

The minimum kinetic energy of the proton.

(f)

Expert Solution
Check Mark

Answer to Problem 38P

The minimum kinetic energy of the proton is 1.88 MeV.

Explanation of Solution

Formula to calculate the minimum kinetic energy of the proton is,

KEmin=(1+m(H11)m(L37i))|Q|

Substitute 1.007825 u for m(H11) , 7.016004 u for m(L37i) and 1.64MeV for Q .

KEmin=(1+1.007825 u7.016004 u)|1.64MeV|=(1+1.007825 u7.016004 u)(1.64MeV)=1.88MeV

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

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