The threshold wavelength for photoemission.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 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 planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =

simple diagram to illustrate the setup for each law- coulombs law and biot savart law

Answer 2

Question

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 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!

Answer 3

Question

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 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!

Answer 4

Question

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 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!

Answer 5

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 eV_.

Answer 6

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

Answer 7

Chapter 26, Problem 27E

(a)

To determine

The threshold wavelength for photoemission.

Answer 8

(a)

Expert Solution

Answer to Problem 27E

Thethreshold wavelength is 276 nm_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given that the work function is 4.49 eV.

Write the expression for the work function in terms of the threshold wavelength.

W=hcλ0 (I)

Here, W is the work function is, h is the Planck’s constant, c is the speed of light, and λ0 is the threshold wavelength.

Solve Equation (I) for λ0.

λ0=hcW (II)

Conclusion:

Substitute 6.62×10−34 J⋅s for h, 3×108 m/s for c, and 4.49 eV for W in Equation (II) to find the threshold wavelength.

λ0=(6.62×10−34 J⋅s)(3×108 m/s)(4.49 eV×1.6×10−19 J1 eV)=2.76×10−7 m=276 nm

Therefore, the threshold wavelengthis 276 nm_.

Answer 13

(b)

To determine

The maximum kinetic energy of the emitted electrons.

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

Answer 14

(b)

To determine

The maximum kinetic energy of the emitted electrons.

Answer 15

(b)

Expert Solution

Answer to Problem 27E

The maximum kinetic energy of the emitted electrons is 0.47 eV_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given that the wavelength of ultraviolet light is 250 nm.

Write the expression for the maximum kinetic energy of the ejected electrons.

Kmax=E−W (III)

Here, KEmax is the maximum kinetic energy with which the electrons are ejected, and E is the energy of the incident photons.

Write the expression relating energy, wavelength, speed of light and Planck’s constant.

E=hcλ (IV)

Here, λ is the wavelength of the photon.

Use Equation (IV) in (III).

Kmax=hcλ−W (V)

Conclusion:

Substitute 4.1357×10−15 eV⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W in Equation (V) to find the maximum kinetic energy of the emitted electrons.

Kmax=(4.1357×10−15 eV⋅s)(3×108 m/s)250 nm×10−9 m1 nm−(4.49 eV)=0.47 eV

Therefore, the maximum kinetic energy of the emitted electrons is 0.47 eV_.

Answer 20

(c)

To determine

The stopping potential.

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 eV_.

Answer 21

(c)

To determine

The stopping potential.

Answer 22

(c)

Expert Solution

Answer to Problem 27E

The stopping potentialis 0.48 eV_.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for Einstein’s photoelectric equation.

eV0=hf−W (VI)

Here, e is the charge of electron, V0 is the stopping potential, and f is the frequency of the incident light.

Write the expression for frequency.

f=cλ (VII)

Use Equation (VII) in (VI) and solve for V0.

eV0=hcλ−WV0=1e(hcλ−W) (VIII)

Conclusion:

Substitute 1.6×10−19 C for e, 6.62×10−34 J⋅s for h, 3×108 m/s for c, 250 nm for λ, and 4.49 eV for W.

V0=1(1.6×10−19 C)((6.62×10−34 J⋅s)(3×108 m/s)250 nm×10−9 m1 nm−((4.49 eV)(1.6×10−19 J1 eV)))=0.475 eV≈0.48 eV

Therefore, the stopping potentialis 0.48 eV_.

Answer 27

Ch. 26 - Prob. 1RQ Ch. 26 - Prob. 2RQ Ch. 26 - Prob. 3RQ Ch. 26 - Prob. 4RQ Ch. 26 - Prob. 5RQ Ch. 26 - Prob. 6RQ Ch. 26 - Prob. 7RQ Ch. 26 - Prob. 1E Ch. 26 - Prob. 2E Ch. 26 - Prob. 3E

The threshold wavelength for photoemission.

Concept explainers

The threshold wavelength for photoemission.

Answer to Problem 27E

Explanation of Solution

The maximum kinetic energy of the emitted electrons.

Answer to Problem 27E

Explanation of Solution

The stopping potential.

Answer to Problem 27E

Explanation of Solution

Want to see more full solutions like this?

Chapter 26 Solutions