The energy and momentum of one laser photon if the laser wavelength is 670 nm .

Question

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Answer 1

Question

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

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Answer 2

Question

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

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Answer 3

Question

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

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Answer 4

Question

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

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Answer 5

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Answer 6

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Answer 7

Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670 nm$ .

Answer 8

(a)

Expert Solution

Answer to Problem 93P

The energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation for the energy of the photon.

$E=hcλ$ (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $λ$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

$E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

$p=Ec$ (II)

Conclusion:

The value of $hc$ is $1240 eV⋅nm$ and the value of $c$ is $3.00×108 m/s$ .

Substitute $1240 eV⋅nm$ for $hc$ and $670 nm$ for $λ$ in equation (I) to find $E$ .

$E=1240 eV⋅nm670 nm=1.85 eV≈1.9 eV$

Substitute $1.85 eV$ for $E$ and $3.00×108 m/s$ for $c$ in equation (II) to find $p$ .

$p=1.85 eV1.6×10−19 J1 eV3.00×108 m/s=9.9×10−28 kg⋅m/s$

Therefore, the energy of the photon is $1.9 eV$ and its momentum is $9.9×10−28 kg⋅m/s$ .

Answer 13

(b)

To determine

The number of photons per second emitted by the laser.

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

Answer 14

(b)

To determine

The number of photons per second emitted by the laser.

Answer 15

(b)

Expert Solution

Answer to Problem 93P

The number of photons per second emitted by the laser is $3×1015 photons/s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

$rate=PE$ (III)

Conclusion:

Given that the output power of the laser pointer is $1 mW$ .

Substitute $1 mW$ for $P$ and $1.85 eV$ for $E$ in equation (III) to find the photon emission rate.

$rate=(1 mW1 W103 mW)(1.85 eV1.6×10−19 J1 eV)=1×10−3 W2.96×10−19 J=3×1015 photons/s$

Therefore, number of photons per second emitted by the laser is $3×1015 photons/s$ .

Answer 20

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)

Conclusion:

Substitute $1 mW$ for $P$ and $3.00×108 m/s$ for $c$ in equation (VII) to find $Fav$ .

$Fav=(1 mW1 W103 mW)3.00×108 m/s=1×10−3 W3.00×108 m/s=3×10−12 N$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Answer 21

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

Answer 22

(c)

Expert Solution

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3×10−12 N$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

$Δp=FavΔt$

Here, $Δp$ is the change in momentum, $Fav$ is the average force applied and $Δt$ is the time interval.

Rewrite the above equation for $Fav$ .

$Fav=ΔpΔt$ (IV)

Use equation (II) to write the expression for the change in momentum.

$Δp=ΔEc$ (V)

Write the equation for $Δt$ .

$Δt=ΔEP$ (VI)

Put equations (V) and (VI) in equation (IV).

$Fav=ΔE/cΔE/P=Pc$ (VII)