The mass of the particles on the left side.

Question

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Answer 1

Question

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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

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 $m(n01)$ .

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

Conclusion:

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

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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

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

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

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Answer 2

Question

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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

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 $m(n01)$ .

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

Conclusion:

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

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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

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

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

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Answer 3

Question

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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

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 $m(n01)$ .

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

Conclusion:

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

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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

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

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

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Answer 4

Question

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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

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 $m(n01)$ .

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

Conclusion:

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

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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

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

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

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Answer 5

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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

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 $m(n01)$ .

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

Conclusion:

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

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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

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

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

Answer 6

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

(a)

Expert Solution

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.

Answer 7

Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

Answer 8

(a)

Expert Solution

Answer to Problem 38P

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

Answer 13

(b)

To determine

The mass of the particles on the right side.

(b)

Expert Solution

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 $m(n01)$ .

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

Conclusion:

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

Answer 14

(b)

To determine

The mass of the particles on the right side.

Answer 15

(b)

Expert Solution

Answer to Problem 38P

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

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 $m(n01)$ .

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

Conclusion:

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

Answer 20

(c)

To determine

The Q-value.

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

Answer 21

(c)

To determine

The Q-value.

Answer 22

(c)

Expert Solution

Answer to Problem 38P

The Q-value is

$−1.64 MeV$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula to calculate the Q-value is,

$Q=(mi−mf)c2$

c is the speed of light.

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

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

Conclusion:

The Q-value is $−1.64 MeV$ .

Answer 27

(d)

To determine

The expression describing the law of conservation of momentum.

(d)

Expert Solution

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 $mpv=(mBe+mn)V$

Answer 28

(d)

To determine

The expression describing the law of conservation of momentum.

Answer 29

(d)

Expert Solution

Answer to Problem 38P

The expression describing the law of conservation of momentum is

$mpv=(mBe+mn)V$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

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 $mpv=(mBe+mn)V$

Answer 34

(e)

To determine

The expression relating the kinetic energies of the particles.

(e)

Expert Solution

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

Answer 35

(e)

To determine

The expression relating the kinetic energies of the particles.

Answer 36

(e)

Expert Solution

Answer to Problem 38P

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

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

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

Answer 41

(f)

To determine

The minimum kinetic energy of the proton.

(f)

Expert Solution

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

Answer 42

(f)

To determine

The minimum kinetic energy of the proton.

Answer 43

(f)

Expert Solution

Answer to Problem 38P

The minimum kinetic energy of the proton is 1.88 MeV.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

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 $−1.64 MeV$ for $Q$ .

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

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV

Answer 48

Ch. 29.3 - Prob. 29.1QQ Ch. 29.3 - What fraction of a radioactive sample has decayed...Ch. 29.3 - Prob. 29.3QQ Ch. 29.6 - Prob. 29.4QQ Ch. 29.6 - Prob. 29.5QQ Ch. 29 - Prob. 1CQ Ch. 29 - Prob. 2CQ Ch. 29 - Prob. 3CQ Ch. 29 - Prob. 4CQ Ch. 29 - Prob. 5CQ

The mass of the particles on the left side.

Concept explainers

The mass of the particles on the left side.

Answer to Problem 38P

Explanation of Solution

The mass of the particles on the right side.

Answer to Problem 38P

Explanation of Solution

The Q-value.

Answer to Problem 38P

Explanation of Solution

The expression describing the law of conservation of momentum.

Answer to Problem 38P

Explanation of Solution

The expression relating the kinetic energies of the particles.

Answer to Problem 38P

Explanation of Solution

The minimum kinetic energy of the proton.

Answer to Problem 38P

Explanation of Solution

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