The wavelengths of photons scattered at angles 30.0 ° , 60.0 ° , 90.0 ° , 120.0 ° , 150.0 ° , 180.0 ° and 210 ° .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 3, Problem 26P

(a)

To determine

The wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210°.

(a)

Expert Solution

Answer to Problem 26P

The wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 4.03×10−11 m, 4.12×10−11 m, 4.24×10−11 m, 4.36×10−11 m, 4.45×10−11 m 4.48×10−11 m and 4.45×10−11 m respectively.

Explanation of Solution

Write the expression to calculate the change in wavelength.

Δλ=hmec(1−cosθ) (I)

Here, h is the Plank’s constant, me is the mass of electron, c is the speed of light, Δλ is the change in wavelength, θ is the scattering angle.

Write the expression to calculate the wavelengths of scattered photons.

λ=λ′+Δλ (II)

Here, λ is the wavelengths of scattered photons, λ′ is the wavelength of X-rays before scattering.

Substitute equation (I) in (II) to calculate the wavelength of scattered photons.

λ=λ′+hmec(1−cosθ) (III)

Conclusion:

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 30.0° for θ to find the wavelength of scattered photon at 30.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(30.0°))→=0.040×10−9 m+0.3×10−12 m=4.03×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 60.0° for θ to find the wavelength of scattered photon at 60.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(60.0°))=0.040×10−9 m+1.2×10−12 m=4.12×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 90.0° for θ to find the wavelength of scattered photon at 90.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(90.0°))=0.040×10−9 m+2.4×10−12 m=4.24×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 120.0° for θ to find the wavelength of scattered photon at 120.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(120.0°))=0.040×10−9 m+3.6×10−12 m=4.36×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 150.0° for θ to find the wavelength of scattered photon at 150.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(150.0°))=0.040×10−9 m+4.5×10−12 m=4.45×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c and 180.0° for θ to find the wavelength of scattered photon at 180.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(180.0°))=0.040×10−9 m+4.8×10−12 m=4.48×10−11 m

Substitute 0.040nm for λ′, 6.626×10−34 J⋅s for h, 9.11×10−31 kg for me, 2.998×108 m/s for c

and 210.0° for θ to find the wavelength of scattered photon at 180.0° in expression (III).

λ=0.040nm(10−9 m1.0 nm)+6.626×10−34 J⋅s(9.11×10−31 kg)(2.998×108 m/s)(1−cos(210.0°))=0.040×10−9 m+4.5×10−12 m=4.45×10−11 m

Therefore the wavelengths of photons scattered at angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 4.03×10−11 m, 4.12×10−11 m, 4.24×10−11 m, 4.36×10−11 m, 4.45×10−11 m 4.48×10−11 m and 4.45×10−11 m respectively.

(b)

To determine

The energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210°.

(b)

Expert Solution

Answer to Problem 26P

The energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 231 eV, 905 eV, 1760eV, 2570 eV, 3140 eV, 3330 eV and 3140 eV respectively.

Explanation of Solution

The energy of the electrons will be the energy lost from the photons.

Write the expression to find the energy of electrons.

Ke=hcλ0−hcλ′=hc(1λ′−1λ) (I)

Here, Ke is the energy of electrons, λ′ is the wavelength before scattering, λ is the wavelength of scattered photon.

Conclusion:

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0403×10−9 m for λ to find the energy of scattered electron when scattering angle is 30.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0403×10−9 m)(1.0 J1.602×10−19 eV)=231 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0412×10−9 m for λ to find the energy of scattered electron when scattering angle at 60.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0412×10−9 m)(1.0 J1.602×10−19 eV)=905 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0424×10−9 m for λ to find the energy of scattered electron when scattering angle at 90.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0424×10−9 m)(1.0 J1.602×10−19 eV)=1760 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0436×10−9 m for λ to find the energy of scattered electron when scattering angle at 120.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0436×10−9 m)(1.0 J1.602×10−19 eV)=2570 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0445×10−9 m for λ to find the energy of scattered electron when scattering angle at 150.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0445×10−9 m)(1.0 J1.602×10−19 eV)=3140 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0448×10−9 m for λ to find the energy of scattered electron when scattering angle at 180.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0448×10−9 m)(1.0 J1.602×10−19 eV)=3330 eV

Substitute 0.040×10−9 m for λ′, 6.626×10−34 J⋅s for h, 2.998×108 m/s for c and 0.0445×10−9 m for λ to find the energy of scattered electron when scattering angle at 210.0° in expression (I).

Ke=(6.626×10−34 J⋅s)(2.998×108 m/s)(10.040×10−9 m−10.0445×10−9 m)(1.0 J1.602×10−19 eV)=3140 eV

Therefore, the energy of electrons at scattering angles 30.0°, 60.0°, 90.0° , 120.0°, 150.0°, 180.0° and 210° are 231 eV, 905 eV, 1760eV, 2570 eV, 3140 eV, 3330 eV and 3140 eV respectively.

(c)

To determine

The scattering angle with which the electron gets the greatest energy.

(c)

Expert Solution

Answer to Problem 26P

The scattering angle with which the electron gets the greatest energy is 180°.

Explanation of Solution

When the scattering angle is 180.0°, the photon strikes on the electron which is at rest with a head on collision. In the head on collision, the photon imparts its momentum to the electron.

After the head on collision the photon scattered straight back and the kinetic energy gained by the initially stationary electron will be the maximum. The kinetic energy gained by the electron at scattering angle 180° is 3330 eV.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 24.5° with the horizontal. A 30.0-kg piece of luggage is placed on the carousel, 7.46 m from the axis of rotation. The travel bag goes around once in 37.5 s. Calculate the magnitude of the force of static friction between the bag and the carousel. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to a position 7.94 m from the axis of rotation. The bag is on the verge of slipping as it goes around once every 30.5 s. Calculate the coefficient of static friction between the bag and the carousel. Your response differs significantly from the correct answer. Rework your solution from the…

(a) Imagine that a space probe could be fired as a projectile from the Earth's surface with an initial speed of 5.78 x 104 m/s relative to the Sun. What would its speed be when it is very far from the Earth (in m/s)? Ignore atmospheric friction, the effects of other planets, and the rotation of the Earth. (Consider the mass of the Sun in your calculations.) Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s (b) What If? The speed provided in part (a) is very difficult to achieve technologically. Often, Jupiter is used as a "gravitational slingshot" to increase the speed of a probe to the escape speed from the solar system, which is 1.85 x 104 m/s from a point on Jupiter's orbit around the Sun (if Jupiter is not nearby). If the probe is launched from the Earth's surface at a speed of 4.10 x 10 m/s relative…

As shown in the figure, a roller-coaster track includes a circular loop of radius R in a vertical plane. A car of mass m is released from rest at a height h above the bottom of the circular section and then moves freely along the track with negligible energy loss due to friction. i (a) First suppose the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height h of the release point above the bottom of the loop. (Use any variable or symbol stated above along with the following as necessary: g.) h = (b) If the car is released at some point above the minimum required height, determine the amount by which the normal force on the car at the bottom of the loop exceeds the normal force on the car at the top of the loop. (Consider the moments when the car reaches the top and when it reaches the bottom again. Use any variable or symbol stated above along with the following as necessary: g.) NB - NT = The normal force…

Answer 2