Question 1

What is the temperature of a burner on an electric stove when its glow is barely visible, at a wavelength of 700 nm? Assume the burner radiates as an ideal blackbody and that 700 nm represents the peak of its emission spectrum.

Accepted Answer

The answer of What is the temperature of a burner...

Question 2

Electromagnetic radiation with a wavelength of 3.5 *10^-12 m is scattered from stationary electrons and photons that have been scattered through 50° are detected.An electron from which one of these photons was scattered receives an energy of:

Accepted Answer

The energy of a photon with wavelength λ is given by E = hc/λ, where h is Planck's constant and c is the speed of light. 
For the given wavelength, 
λ = 3.5 x 10^-12 m 
E = hc/λ = (6.626 x 10^-34 J s) (3 x 10^8 m/s) / (3.5 x 10^-12 m) = 1.79 x 10^-19 J 
The energy of the photon scattered through 50 degrees is the same as the incident photon. 
Therefore, the energy received by the electron is the difference between the energies of the incident and scattered photon. 
ΔE = Eincident - Escattered = (1.79 x 10^-19 J) - (1.1 x 10^-14 J) = -1.098 x 10^-14 J 
Since the electron is stationary, it does not receive any kinetic energy from the scattered photon, and the answer is A) 0 J. 
However, this is not one of the options, so the closest option is B) 1.1 x 10^-14 J.

Question 3

Electromagnetic radiation with a wavelength of 5.7 * 10^-12 m is incident on stationary electrons.Radiation that has a wavelength of 6.6 *10^-12 m is detected at a scattering angle of:

Accepted Answer

The answer of Electromagnetic radiation with a wavelength of 5.7...

Question 4

The main problem that physicists had in understanding blackbody radiation before Planck's work was:

Accepted Answer

The main issue was the "ultraviolet catastrophe," where classical physics predicted an infinitely large peak at high frequencies, which did not match experimental observations. Max Planck resolved this by introducing the concept of quantized energy levels.

Question 5

The frequency and wavelength of the matter wave associated with a 10-eV free electron are:

Accepted Answer

The frequency and wavelength of a matter wave can be calculated using the de Broglie wavelength formula, $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. For a free electron with a given energy, its momentum can be derived from its kinetic energy, and subsequently, its frequency can be calculated using the relation $f = \frac{E}{h}$, where $E$ is the energy of the electron. The correct values align with the principles of quantum mechanics and the calculations for a 10-eV electron.

Question 6

In Compton scattering from stationary particles the maximum change in wavelength can be made smaller by using:

Accepted Answer

The maximum change in wavelength in Compton scattering is inversely proportional to the mass of the particle. Using more massive particles reduces the maximum change in wavelength.

Question 7

Consider the following: I.A photoelectric process in which all emitted electrons have energy less than hf, where f is the frequency of the incident light.
II.A photoelectric process in which some emitted electrons have kinetic energy greater than hf.
III.Compton scattering from stationary electrons for which the emitted light has a frequency that is greater than that of the incident light.
IV.Compton scattering from stationary electrons for which the emitted light has a frequency that is less than that of the incident light.
The only possible processes are:

Accepted Answer

Process I is a standard photoelectric effect where all emitted electrons have energy less than the energy of the incident photons. Process II is a photoelectric effect where some electrons are emitted with higher energy than the incident photons, which is not possible according to the laws of conservation of energy. Process III is Compton scattering from stationary electrons where the frequency of the emitted photon is greater than that of the incident photon. Finally, process IV is Compton scattering from stationary electrons where the frequency of the emitted photon is less than that of the incident photon, which is possible due to the conservation of momentum and energy. Therefore, only processes I and IV are possible.

Question 8

Monoenergetic electrons are incident on a single slit barrier.If the energy of each incident electron is increased the central maximum of the diffraction pattern:

Accepted Answer

The width of the central maximum in the diffraction pattern is dependent on the wavelength of the incident electrons, which is inversely proportional to their energy. As the energy of the electrons is increased, their wavelength decreases, and therefore the central maximum narrows.

Question 9

The surface of the Sun is at a temperature of approximately 5800 K, and radiates a peak wavelength of 500 nm.According to the Planck radiation law, what is its emitted intensity per unit wavelength at the peak?

Accepted Answer

The Planck radiation law is given by:$$I(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_BT} - 1}$$where:* $I(\lambda, T)$ is the emitted intensity per unit wavelength at wavelength $\lambda$ and temperature $T$* $h$ is the Planck constant* $c$ is the speed of light* $k_B$ is the Boltzmann constantAt the peak wavelength, the derivative of $I(\lambda, T)$ with respect to $\lambda$ is zero. This gives:$$\frac{dI(\lambda, T)}{d\lambda} = 0 = \frac{2hc^2}{\lambda^6} \frac{5hc}{\lambda k_BT e^{hc/\lambda k_BT} - 1} - \frac{2hc^2}{\lambda^5} \frac{hc}{\lambda k_BT e^{hc/\lambda k_BT} - 1}$$Solving for $\lambda$, we get:$$\lambda_{	ext{peak}} = \frac{hc}{5k_BT}$$Substituting the given values, we get:$$\lambda_{	ext{peak}} = \frac{(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})}{5(1.381 	imes 10^{-23} 	ext{ J/K})(5800 	ext{ K})}$$$$\lambda_{	ext{peak}} = 500 	ext{ nm}$$Now, we can calculate the emitted intensity per unit wavelength at the peak:$$I(\lambda_{	ext{peak}}, T) = \frac{2hc^2}{\lambda_{	ext{peak}}^5} \frac{1}{e^{hc/\lambda_{	ext{peak}} k_BT} - 1}$$$$I(\lambda_{	ext{peak}}, T) = \frac{2(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})^2}{(500 	imes 10^{-9} 	ext{ m})^5} \frac{1}{e^{(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s})/(500 	imes 10^{-9} 	ext{ m})(1.381 	imes 10^{-23} 	ext{ J/K})(5800 	ext{ K})} - 1}$$$$I(\lambda_{	ext{peak}}, T) = 8.4 	ext{ W/cm}^2\cdot	ext{nm}$$Therefore, the correct answer is A.

Question 10

A free electron has a momentum of 5.0*10^-24kg . m/s.Its wavelength, as given by its wave function, is:

Accepted Answer

The de Broglie wavelength of a particle is given by λ=h/p, where h is Planck's constant and p is the momentum. Plugging in the values given in the problem, we get λ=(6.626×10^-34)/(5.0×10^-24)=1.3×10^-10m. Therefore, the correct answer is B.

Question 11

Electromagnetic radiation with a wavelength of 3.5 * 10^-12 m is scattered from stationary electrons, and photons that have been scattered through 50° are detected.After a scattering event the magnitude of the photon's momentum is:

Accepted Answer

The momentum of a photon is given by p = h/λ, where h is Planck's constant (6.626 x 10^-34 Js) and λ is the wavelength. Substituting the given wavelength (3.5 x 10^-12 m) into the formula gives a momentum of approximately 1.5 x 10^-22 kg·m/s.

Question 12

Consider the following three particles:  Rank them according to the wavelengths of their waves, least to greatest.

Accepted Answer

The answer of Consider the following three particles:  Rank...

Question 13

Of the following which is the best evidence for the wave nature of matter?

Accepted Answer

The reflection of electrons by crystals (also known as electron diffraction) is the best evidence for the wave nature of matter. This phenomenon demonstrates that electrons can exhibit diffraction patterns similar to those of waves, indicating that they have wave-like properties. The other options listed are not as direct in their evidence for the wave nature of matter.

Question 14

Consider the following three particles:  Rank them according to the wavelengths of their matter waves, least to greatest.

Accepted Answer

The wavelength of a matter wave is given by the de Broglie wavelength formula: 
λ = h/p
where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. 
Higher momentum results in smaller wavelengths.

Particle 1: 
No information given about the momentum or mass of the particle. 
Cannot compare its wavelength to the other particles.

Particle 2: 
Mass: 6.64 x 10^-27 kg 
Speed: 4 x 10^4 m/s 
Momentum: p = mv = (6.64 x 10^-27 kg)(4 x 10^4 m/s) = 2.656 x 10^-22 kg·m/s 
λ = h/p = (6.626 x 10^-34 J·s)/(2.656 x 10^-22 kg·m/s) ≈ 2.5 x 10^-12 m

Particle 3: 
Mass: 9.11 x 10^-31 kg 
Speed: 2 x 10^7 m/s 
Momentum: p = mv = (9.11 x 10^-31 kg)(2 x 10^7 m/s) = 1.822 x 10^-23 kg·m/s 
λ = h/p = (6.626 x 10^-34 J·s)/(1.822 x 10^-23 kg·m/s) ≈ 3.6 x 10^-11 m

Since particle 3 has higher momentum than particle 2, its wavelength is smaller. Therefore, we have Particle 2 > Particle 3. 
And we cannot compare the wavelength of Particle 1 with either particles 2 or 3.

Therefore, the correct ranking is 3, 2, 1 or choice B.

Question 15

In Compton scattering from stationary electrons the largest change in wavelength that can occur is:

Accepted Answer

The largest change in wavelength in Compton scattering is given by the Compton wavelength shift formula, which is 2.43 x 10^-12 m for a photon scattering at 180 degrees.

Question 16

Evidence for the wave nature of matter is:

Accepted Answer

The wave nature of matter is evidenced by electron diffraction experiments, such as those conducted by Davisson and Germer, which demonstrated that electrons can exhibit wave-like behavior.

Question 17

Of the following, Compton scattering from electrons is most easily observed for:

Accepted Answer

Compton scattering is most easily observed with x-rays, as they have a shorter wavelength and higher energy, making them more likely to interact with electrons through scatter.

Question 18

In Compton scattering from stationary electrons the frequency of the emitted light is independent of:

Accepted Answer

In Compton scattering, the frequency of the emitted light depends on the frequency of the incident light, the scattering angle, and indirectly on the recoil speed and energy of the electron, as these factors are related to the change in wavelength (and thus frequency) of the scattered photon. Therefore, none of the given options are correct; the frequency of the emitted light is not independent of any of the listed factors.

Question 19

Separate Compton effect experiments are carried out using visible light and x rays.The scattered radiation is observed at the same scattering angle.For these experiments:

Accepted Answer

The answer of Separate Compton effect experiments are carried out...

Question 20

Which of the following is NOT evidence for the wave nature of matter?

Accepted Answer

The photoelectric effect is evidence for the particle nature of light, not the wave nature of matter. It demonstrates how photons (particles of light) can eject electrons from a metal surface, which supports the particle theory of light rather than the wave theory of matter.

Quiz 38: Photons and Matter Waves