Question 1

The energy of an x-ray photon of wavelength 3.0 × 10^-10 m is A) 1.0 × 10^-16 J. B) 9.9 × 10^-16 J. C) 6.6 × 10^-16 J. D) 2.2 × 10^-24 J. E) 3.2 × 10^-29 J.

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. Substituting the given values, we get:
E = (6.626 × 10^-34 J s) × (3.0 × 10^8 m/s) / (3.0 × 10^-10 m)
E = 6.6 × 10^-16 J
Therefore, the correct answer is C, 6.6 × 10^-16 J.

Question 2

Microwaves range in wavelength from about 2.0 × 10^-2 cm to about 5.0 cm. What is the maximum energy in eV that a microwave may have? A) 0.025 meV B) 6.2 meV C) 2.5 eV D) 6.2 eV E) 0.62 meV

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the maximum wavelength of 5.0 cm, we get E = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (5.0 x 10^-2 m) = 3.98 x 10^-23 J. Converting to electron volts (eV), we divide by the elementary charge (1.602 x 10^-19 C/eV) to get 0.025 eV. However, the question asks for the maximum energy, so we need to use the minimum wavelength of 2.0 x 10^-2 cm instead. Plugging that in, we get E = (6.626 x 10^-34 J s) x (2.998 x 10^8 m/s) / (2.0 x 10^-2 m) = 9.94 x 10^-22 J, which is 6.2 meV when converted to eV. Therefore, the correct answer is B.

Question 3

The maximum kinetic energy of photoelectrons produced in the photoelectric effect depends directly on the
A) frequency of the incident photons.
B) intensity of the incident photons.
C) area of the metal surface from which the photoelectrons are released.
D) thickness of the metal.
E) photoelectric current.

Accepted Answer

The maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident photons. This is because the energy of a photon is directly proportional to its frequency (E=hf) and the maximum kinetic energy of a photoelectron is equal to the energy of the incident photon minus the work function of the metal. As such, increasing the frequency of the incident photons increases their energy and therefore increases the maximum kinetic energy of the photoelectrons. The intensity of the incident photons, area of the metal surface, thickness of the metal, and photoelectric current all do not directly affect the maximum kinetic energy of the photoelectrons.

Question 4

The graph shows the stopping potential for photoelectrons as a function of the frequency of the incident light for two different metals Cs and W). Which of the following statements is true?
A) Because the intercepts are different, Planck's constant is not constant.
B) For a given frequency of incident light, the photoelectrons from W are more energetic than those from Cs.
C) The maximum kinetic energy of the photoelectrons increases as the frequency of the incident light increases.
D) The stopping potentials for all metals are the same.
E) The threshold frequency for Cs is greater than that for W.

Accepted Answer

The answer of  The graph shows the stopping potential...

Question 5

This is an apparatus for studying the photoelectric effect. The work function of the material being investigated is 3.8 * 10^-19 J. Light of a wavelength 350 nm is incident on the material. What is the lowest voltage needed between the cathode C) and the anode A) to stop any electrons ejected from the cathode from reaching the anode? A) 2.37 V B) 1.17 V C) 3.54 V D) 4.71 V E) None of these is correct.

Accepted Answer

The answer of  This is an apparatus for studying...

Question 6

The dissociation energy is the energy required to separate the two atoms in a diatomic molecule in their ground state. If the dissociation energy of molecular oxygen is 7.2 eV, then calculate the maximum wavelength of light from the Sun that can break apart atmospheric O₂. A) 1.7 × 10^-7 nm B) 1.7 × 10^-8 m C) 1.7 × 10^-9 m D) 1.7 × 10^-7 m E) 1.7 × 10² m

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant and c is the speed of light. Solving for λ gives λ = hc/E. Substituting h = 4.135667696 × 10^-15 eV·s, c = 3 × 10^8 m/s, and E = 7.2 eV, we find λ ≈ 1.7 × 10^-7 m.

Question 7

Magnetic resonance imaging MRI) is a much-used noninvasive diagnostic technique in medicine. The radio frequency photons used have energies of about 10^-7 eV. What wavelength does this correspond to? A) 12 nm B) 1.2 m C) 12 m D) 1.2 nm E) 120 m

Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant and c is the speed of light. Solving for λ with E = 10^-7 eV gives a wavelength of approximately 12 meters.

Question 8

The energy of a quantum of radiation that has a wavelength of 447 nm in a vacuum is approximately A) 2.37 × 10^-19J. B) 1.60 × 10^-19 J. C) 9.88 × 10^-33 J. D) 9.11 × 10^-31 J. E) None of these is correct.

Accepted Answer

The answer of The energy of a quantum of radiation...

Question 9

The graph shows the maximum kinetic energy of electrons emitted by a photosensitive surface as a function of the frequency of the incident radiation. The slope of this curve represents
A) the intensity of the incident radiation.
B) the maximum kinetic energy.
C) the threshold frequency.
D) Planck's constant.
E) the stopping potential.

Accepted Answer

The slope of this graph represents Planck's constant, denoted by the symbol h. Planck's constant is a fundamental constant of nature that relates the energy of a photon to its frequency. The equation that relates the maximum kinetic energy of electrons emitted by a photosensitive surface to the frequency of the incident radiation involves Planck's constant, which allows us to determine it from the slope of the graph.

Question 10

The photoelectric threshold of a certain metal is 310 nm. The maximum kinetic energy of electrons ejected from the surface of the metal by ultraviolet light of wavelength 200 nm is
A) 2.22 eV.
B) 2.60 eV.
C) 10.2 eV.
D)12.3 eV.
E) None of these because no electrons are ejected.

Accepted Answer

The energy of the incident light (200 nm) is greater than the work function (310 nm), so electrons are ejected. Using the photoelectric equation, the maximum kinetic energy is calculated as 2.22 eV.

Question 11

Light falling on the surface of a metal such as cesium can liberate electrons from the metal. The kinetic energy of electrons emitted from a metal can be increased by
A) using light of higher frequency.
B) using light of lower frequency.
C) increasing the intensity of the incident light.
D) using a metal with a greater work function.
E) None of these is correct.

Accepted Answer

The kinetic energy of electrons emitted from a metal is directly proportional to the frequency of the incident light. Therefore, using light of higher frequency will increase the kinetic energy of the emitted electrons.

Question 12

When a surface is illuminated with light of wavelength 500 nm, the maximum speed of the photoelectrons is 300 km/s. When the same surface is illuminated by light of half this wavelength, what is the maximum speed of the photoelectrons in this case?
A) 680 km/s
B) 1280 km/s
C) 885 km/s
D) 980 km/s
E) 14800 km/s

Accepted Answer

The maximum speed of photoelectrons is given by:

$(1/2)mv^2 = hf - \phi$

where, $m$ is the mass of the electron, $v$ is its speed, $h$ is Planck's constant, $f$ is the frequency of the light, and $\phi$ is the work function of the metal surface.

We can rearrange this equation as:

$v = \sqrt{(2/h)(hf - \phi)}$

We are given that for light of wavelength 500 nm, $v_{max} = 300$ km/s. The frequency of this light is:

$f_1 = c/\lambda_1 = (3 	imes 10^8 m/s)/(500 	imes 10^{-9} m) = 6 	imes 10^{14} Hz$

Using this frequency and $v_{max}$ in the above equation, we can find $\phi$:

$\phi = hf_1 - (1/2)mv_{max}^2 = (6.626 	imes 10^{-34} J s)(6 	imes 10^{14} Hz) - (1/2)(9.11 	imes 10^{-31} kg)(3 	imes 10^5 m/s)^2 = 4.35 	imes 10^{-19} J$

Now, when the same surface is illuminated by light of half the wavelength (250 nm), the frequency of the light is:

$f_2 = c/\lambda_2 = (3 	imes 10^8 m/s)/(250 	imes 10^{-9} m) = 1.2 	imes 10^{15} Hz$

Using this frequency and $\phi$ in the above equation, we can find the new maximum speed of the photoelectrons:

$v = \sqrt{(2/h)(hf_2 - \phi)} = \sqrt{(2/h)(1.2 	imes 10^{15} Hz)(4.35 	imes 10^{-19} J) - 4.35 	imes 10^{-19} J)} = 979.8$ km/s

Therefore, the answer is D) 980 km/s.

Question 13

A Compton-scattered x-ray photon has less energy than the incident photon. The scattered photon therefore
A) has the same wavelength as the incident photon.
B) has a longer wavelength than the incident photon.
C) has a shorter wavelength than the incident photon.
D) cannot be assigned a wavelength.
E) has none of these wavelengths.

Accepted Answer

The Compton effect shows that when an x-ray photon is scattered, it loses energy, resulting in an increase in wavelength according to the equation λ' - λ = (h/mc)(1 - cos θ), where λ' is the wavelength of the scattered photon, λ is the initial wavelength, h is Planck's constant, m is the electron mass, c is the speed of light, and θ is the scattering angle.

Question 14

The work function for a metal is 5.58 eV. What is the threshold wavelength for the photoelectric effect of incident photons? A) 2.22 × 10^-7 m B) 2.22 × 10^-9 m C) 4.50 × 10⁶ m D) 4.50 × 10^-6 m E) 2.22 × 10^-11 m

Accepted Answer

The threshold wavelength is calculated using the formula λ = hc/Φ, where h is Planck's constant, c is the speed of light, and Φ is the work function. Substituting the values, λ = (6.626 × 10^-34 Js)(3 × 10^8 m/s) / (5.58 eV × 1.602 × 10^-19 J/eV) gives approximately 2.22 × 10^-7 m.

Question 15

If the work function of thoriated tungsten is 4 *10^-19 J, the longest wavelength of light that will cause photoelectrons to be emitted is approximately A) 880 nm. B) 400 nm. C) 497 nm. D) 700 nm. E) 181 nm.

Accepted Answer

The answer of If the work function of thoriated tungsten...

Question 16

The graph shows the maximum kinetic energy of electrons emitted from a metal as a function of the frequency of the incident light. The work function of the metal is A) 6.0 eV. B) 4.0 eV. C) 2.5 eV. D) 0.4 eV. E) 4 × 10^-19 eV.

Accepted Answer

The work function is the minimum amount of energy needed to remove an electron from the metal surface. In this graph, the x-intercept represents the threshold frequency below which no electrons are emitted. The threshold frequency corresponds to the threshold energy or the work function. 
From the graph, the threshold frequency is approximately f = 6.5 x 10^14 Hz, which corresponds to a threshold energy of about 2.5 eV. Therefore, the work function of the metal is 2.5 eV. 
The other answer choices can be eliminated because they do not correspond to the threshold energy obtained from the graph.

Question 17

The energy of a photon of visible light is on the order of A) 10^-6 eV. B) 10^-3 eV. C) 1 eV. D) 10³ eV. E) 10⁶ eV.

Accepted Answer

The energy of a photon of visible light is typically around 1 eV, which falls within the range of energies for visible light photons.

Question 18

The pupil area of the human eye is ~ 1 × 10^-5 m². The minimum light intensity that the eye is sensitive to is ~ 1 × 10^-10 Wm^-2. How many photons per second of wavelength 550 nm does this correspond to? A) ~300 s^-1 B) ~30000 s^-1 C) ~3000 s^-1 D) ~3 × 10^-2 s^-1 E) ~3 × 10¹² s^-1

Accepted Answer

The energy of a single photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The number of photons per second is then given by P = I/E, where I is the light intensity. Plugging in the given values, we get:P = (1 × 10^-10 Wm^-2) / ((6.63 × 10^-34 Js) / (550 × 10^-9 m)) = 3000 s^-1Therefore, the correct answer is C.

Question 19

Which of the following formulas has the correct units for the linear momentum of a photon?
A) hc/λ
B) c/n
C) l/λ
D) λ /h
E) None of these is correct.

Accepted Answer

The answer of Which of the following formulas has the...

Question 20

An x-ray photon scattering off an electron imparts some of its energy to the electron the Compton effect). Which of the following statements is true?
A) The wavelength of the scattered photon is unchanged.
B) The wavelength of the scattered photon is decreased.
C) The wavelength of the scattered photon is increased.
D) The scattered photon gains speed.
E) The scattered photon is slowed.

Accepted Answer

In the Compton effect, the scattered photon loses energy to the electron, resulting in an increase in the wavelength of the scattered photon.

Quiz 26: Quantum Physics and Atomic Structure

The energy of an x-ray photon of wavelength 3.0 × 10^-10 m is

Microwaves range in wavelength from about 2.0 × 10^-2 cm to about 5.0 cm. What is the maximum energy in eV that a microwave may have?

The maximum kinetic energy of photoelectrons produced in the photoelectric effect depends directly on the

The graph shows the stopping potential for photoelectrons as a function of the frequency of the incident light for two different metals Cs and W) . Which of the following statements is true?

The dissociation energy is the energy required to separate the two atoms in a diatomic molecule in their ground state. If the dissociation energy of molecular oxygen is 7.2 eV, then calculate the maximum wavelength of light from the Sun that can break apart atmospheric O₂.

Magnetic resonance imaging MRI) is a much-used noninvasive diagnostic technique in medicine. The radio frequency photons used have energies of about 10^-7 eV. What wavelength does this correspond to?

The energy of a quantum of radiation that has a wavelength of 447 nm in a vacuum is approximately

The graph shows the maximum kinetic energy of electrons emitted by a photosensitive surface as a function of the frequency of the incident radiation. The slope of this curve represents

The photoelectric threshold of a certain metal is 310 nm. The maximum kinetic energy of electrons ejected from the surface of the metal by ultraviolet light of wavelength 200 nm is

Light falling on the surface of a metal such as cesium can liberate electrons from the metal. The kinetic energy of electrons emitted from a metal can be increased by

When a surface is illuminated with light of wavelength 500 nm, the maximum speed of the photoelectrons is 300 km/s. When the same surface is illuminated by light of half this wavelength, what is the maximum speed of the photoelectrons in this case?

A Compton-scattered x-ray photon has less energy than the incident photon. The scattered photon therefore

The work function for a metal is 5.58 eV. What is the threshold wavelength for the photoelectric effect of incident photons?

If the work function of thoriated tungsten is 4 *10^-19 J, the longest wavelength of light that will cause photoelectrons to be emitted is approximately

The graph shows the maximum kinetic energy of electrons emitted from a metal as a function of the frequency of the incident light. The work function of the metal is

The energy of a photon of visible light is on the order of

The pupil area of the human eye is ~ 1 × 10^-5 m². The minimum light intensity that the eye is sensitive to is ~ 1 × 10^-10 Wm^-2. How many photons per second of wavelength 550 nm does this correspond to?

Which of the following formulas has the correct units for the linear momentum of a photon?

An x-ray photon scattering off an electron imparts some of its energy to the electron the Compton effect) . Which of the following statements is true?