Question 1

An 84-kW AM radio station broadcasts at 1000 kHz. How many photons are emitted each second by the transmitting antenna? (h = 6.626 × 10^-34 J ∙ s) A) 1.3 × 10³² B) 2.9 × 10²⁴ C) 6.3 × 10¹² D) 1.4 × 10¹⁵

Accepted Answer

The energy of one photon is given by E = hν, where ν is the frequency. For 1000 kHz, ν = 1000 × 10^3 Hz. The energy of one photon is E = 6.626 × 10^-34 J ∙ s × 1000 × 10^3 Hz = 6.626 × 10^-28 J. The number of photons emitted per second is the total power divided by the energy per photon: 84,000 W / 6.626 × 10^-28 J = 1.3 × 10^32 photons per second.

Question 2

If the accuracy in measuring the velocity of a particle increases, the accuracy in measuring its position will
A) increase.
B) decrease.
C) remain the same.
D) It is impossible to say since the two measurements are independent and do not affect each other.

Accepted Answer

According to the Heisenberg uncertainty principle, there is a trade-off between the accuracy in measuring the position and velocity of a particle. The more accurately we know the velocity of a particle, the less accurately we can know its position. Therefore, if the accuracy in measuring the velocity increases, the accuracy in measuring its position will decrease.

Question 3

A photon of initial wavelength 0.651 nm, after being scattered from a free electron at rest, moves off at an angle of 120° with respect to its incident direction. (mel = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸m/s) (a) What is the wavelength of the scattered photon? (b) What is the energy of the scattered photon?

Accepted Answer

(a) 0.655 nm (b) 3.04 × 10^-16J

Question 4

When a metal surface is illuminated with light of wavelength 437 nm, the stopping potential for photoelectrons is 1.67 V. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, e = - 1.60 × 10^-19 C, 1 eV = 1.60 × 10^-19 J, mel = 9.11 × 10^-31 kg) (a) What is the work function of the metal, in eV? (b) What is the maximum speed of the ejected electrons?

Accepted Answer

The answer of When a metal surface is illuminated with...

Question 5

A metal having a work function of 2.5 eV is illuminated with white light that has a continuous wavelength band from 400 nm to 700 nm. For which one of the following ranges of the wavelength band in this white light are photoelectrons NOT produced? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J) A) 500 nm to 700 nm B) 400 nm to 560 nm C) 500 nm to 560 nm D) 400 nm to 500 nm E) 560 nm to 700 nm

Accepted Answer

The answer of A metal having a work function of...

Question 6

When a certain metal is illuminated by light, photoelectrons are observed provided that the wavelength of the light is less than 669 nm. Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J) A) 1.9 eV B) 2.0 eV C) 2.2 eV D) 2.3 eV

Accepted Answer

The work function is given by the equation: hc/λ = Φ + KE, where hc/λ is the energy of the incident photon, Φ is the work function, and KE is the kinetic energy of the photoelectron. Since the photoelectrons are observed when the wavelength of the light is less than 669 nm, the energy of the incident photon is greater than the work function. Therefore, the work function must be less than the energy of the incident photon, which is hc/λ = (6.626 × 10^-34 J s)(3.00 × 10^8 m/s)/(669 × 10^-9 m) = 2.96 eV. The closest value to this is 2.0 eV.

Question 7

A metal having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 4.0 eV. What is the maximum kinetic energy of the photoelectrons produced by this light? (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 2.6 × 10^-19J B) 3.8 × 10^-19 J C) 4.7 × 10^-19J D) 5.5 × 10^-19J E) 6.4 × 10^-19 J

Accepted Answer

The maximum kinetic energy of the photoelectrons is given by the equation: KE_max = Photon energy - Work function. Converting the energies from eV to J: Photon energy = 4.0 eV = 4.0 × 1.60 × 10^-19 J = 6.4 × 10^-19 J, Work function = 2.4 eV = 2.4 × 1.60 × 10^-19 J = 3.84 × 10^-19 J. Therefore, KE_max = 6.4 × 10^-19 J - 3.84 × 10^-19 J = 2.56 × 10^-19 J, which is approximately 2.6 × 10^-19 J.

Question 8

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s = 4.141 × 10^-15 ev ∙ s, 1 eV = 1.60 × 10^-19 J) A) 4.52 eV B) 3.11 eV C) 1.41 eV D) 2.82 eV E) 1.70 eV

Accepted Answer

The energy of the incident photon is calculated using E = hc/λ. Converting 400 nm to meters gives 400 × 10^-9 m. Substituting the values, E = (4.141 × 10^-15 eV ∙ s)(3.00 × 10^8 m/s) / (400 × 10^-9 m) = 3.11 eV. The maximum kinetic energy of the photoelectrons is the photon energy minus the work function: 3.11 eV - 1.70 eV = 1.41 eV.

Question 9

If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will
A) increase.
B) decrease.
C) remain the same.
D) It is impossible to say since the two measurements are independent and do not affect each other.

Accepted Answer

According to the Heisenberg uncertainty principle, there is a trade-off between the accuracy in measuring the position and velocity of a particle. This means that if the position is measured with high accuracy, the velocity measurement will become less accurate. Therefore, as the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will decrease.

Question 10

A metal having a work function of 2.8 eV is illuminated with monochromatic light whose photon energy is 3.9 eV. What is the threshold frequency for photoelectron production? (h = 6.626 × 10^-34J ∙ s, 1 eV = 1.60 × 10^-19J) A) 6.8 × 10¹⁴ Hz B) 2.7 × 10¹⁴ Hz C) 7.6 × 10¹⁴ Hz D) 8.5 × 10¹⁴ Hz E) 9.4 × 10¹⁴ Hz

Accepted Answer

The threshold frequency (f₀) is calculated using the work function (Φ) and Planck's constant (h): f₀ = Φ/h. Convert the work function from eV to joules: 2.8 eV × 1.60 × 10⁻¹⁹ J/eV = 4.48 × 10⁻¹⁹ J. Then, f₀ = 4.48 × 10⁻¹⁹ J / 6.626 × 10⁻³⁴ J∙s = 6.8 × 10¹⁴ Hz.

Question 11

A metal surface has a work function of 1.50 eV. Calculate the maximum kinetic energy, in eV, of electrons ejected from this surface by electromagnetic radiation of wavelength 311 nm. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, e = - 1.60 × 10^-19 C, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of A metal surface has a work function...

Question 12

Gamma rays are photons with very high energy. How many visible-light photons with a wavelength of 500 nm would you need to match the energy of a gamma-ray photon with energy 4.1 × 10^-13 J? (h = 6.626 × 10^-34J ∙ s, c = 3.00 × 10⁸m/s) A) 1.0 × 10⁶ B) 1.4 × 10⁸ C) 6.2 × 10⁹ D) 3.9 × 10³

Accepted Answer

The answer of Gamma rays are photons with very high...

Question 13

In a particular case of Compton scattering, a photon collides with a free electron and scatters backwards. The wavelength after the collision is exactly double the wavelength before the collision. What is the wavelength of the incident photon? (mel = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s) A) 3.6 pm B) 4.8 pm C) 2.4 pm D) 1.2 pm E) 6.0 pm

Accepted Answer

The answer of In a particular case of Compton scattering,...

Question 14

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c = 3.00 × 10⁸m/s, h = 6.626 × 10^-34 J ∙ s) A) 2.8 × 10¹⁸ B) 6.9 × 10¹⁹ C) 3.4 × 10¹⁹ D) 1.1 × 10¹⁷ E) 2.2 × 10¹⁷

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 total energy of the pulse is 1.2 J, so the number of photons emitted is:N = E/(hc/λ) = (1.2 J)/(6.626 × 10^-34 J ∙ s * 3.00 × 10^8 m/s * 463 × 10^-9 m) = 2.8 × 10^18 photons

Question 15

In a photoelectric effect experiment, electrons emerge from a copper surface with a maximum kinetic energy of 1.10 eV when light shines on the surface. The work function of copper is 4.65 eV. Which one of the following values is closest to the wavelength of the light? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J) A) 220 nm B) 150 nm C) 360 nm D) 1100 nm

Accepted Answer

We can use the formula for the photoelectric effect:

Kinetic energy of electron = Energy of photon - Work function

We are given the maximum kinetic energy of the electrons and the work function of copper:

1.10 eV = Energy of photon - 4.65 eV

Solving for the energy of the photon:

Energy of photon = 5.75 eV

We can convert this energy to wavelength using the formula:

Energy of photon = hc/λ

where h is Planck's constant and c is the speed of light. Rearranging the formula:

λ = hc/Energy of photon

Plugging in the values and converting electron volt to joules:

λ = (6.626 × 10^-34 J∙s) × (3.00 × 10^8 m/s) / (5.75 × 1.60 × 10^-19 J)

λ = 220 nm

So the closest choice is A) 220 nm.

Question 16

A beam of red light and a beam of violet light each deliver the same power on a surface. For which beam is the number of photons hitting the surface per second the greatest?
A) the red beam
B) the violet beam
C) It is the same for both beams.

Accepted Answer

The energy of a photon is given by $E = hf$, where $h$ is Planck's constant and $f$ is the frequency of the light. Since violet light has a higher frequency than red light, each photon of violet light carries more energy. Therefore, for two beams delivering the same power, the beam with the lower energy photons (red light) must have a greater number of photons hitting the surface per second.

Question 17

A light beam from a 2.1-mW He-Ne laser has a wavelength of 633 nm. How many photons does the laser emit in one second? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s) A) 6.7 × 10¹⁵ B) 8.8 × 10¹⁵ C) 1.1 × 10¹⁶ D) 1.3 × 10¹⁶

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 emitted per second is given by P/E, where P is the power of the laser. Plugging in the given values, we get:E = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (633 × 10^-9 m) = 3.14 × 10^-19 JP/E = (2.1 × 10^-3 W) / (3.14 × 10^-19 J) = 6.7 × 10^15 photons/s

Question 18

Monochromatic light strikes a metal surface and electrons are ejected from the metal. If the intensity of the light is increased, what will happen to the ejection rate and maximum energy of the electrons?
A) greater ejection rate; same maximum energy
B) same ejection rate; greater maximum energy
C) greater ejection rate; greater maximum energy
D) same ejection rate; same maximum energy

Accepted Answer

Increasing the intensity of monochromatic light increases the number of photons hitting the metal surface per unit time, which increases the rate of electron ejection. However, the maximum energy of the ejected electrons is determined by the frequency of the light, not its intensity, and remains unchanged.

Question 19

A stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 590 nm. Monochromatic light of a different wavelength is now shown on the tube, and the stopping potential is measured to be 2.30 V. What is the wavelength of this new light? (c = 3.00 × 10⁸ m/s, e = - 1.60 × 10^-19 C, h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J) A) 320 nm B) 300 nm C) 340 nm D) 360 nm E) 410 nm

Accepted Answer

The correct answer is A because the energy of the incident photon is equal to the work function of the metal plus the kinetic energy of the emitted electron. The work function is a constant for a given metal, so the kinetic energy of the emitted electron is proportional to the energy of the incident photon. The energy of a photon is inversely proportional to its wavelength, so the shorter the wavelength of the incident light, the greater the kinetic energy of the emitted electron. Therefore, the stopping potential, which is proportional to the kinetic energy of the emitted electron, will be greater for shorter wavelengths of incident light.

Question 20

Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 1.45 eV. What is the work function of this metal? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J) A) 3.73 eV B) 3.13 eV C) 4.33 eV D) 4.92 eV

Accepted Answer

The work function is given by the equation:$$KE_{max} = hν - Φ$$where:* $$KE_{max}$$ is the maximum kinetic energy of the ejected electrons* h is Planck's constant* ν is the frequency of the incident photons* Φ is the work functionWe can rearrange this equation to solve for the work function:$$Φ = hν - KE_{max}$$Plugging in the given values, we get:$$Φ = (6.626 × 10^{-34} J \cdot s)(3.00 × 10^8 m/s)/(240 × 10^{-9} m) - (1.45 eV)(1.60 × 10^{-19} J/eV)$$$$Φ = 3.73 eV$$Therefore, the work function of the metal is 3.73 eV.

Quiz 37: Photons: Light Waves Behaving As Particles

An 84-kW AM radio station broadcasts at 1000 kHz. How many photons are emitted each second by the transmitting antenna? (h = 6.626 × 10-34 J ∙ s)

If the accuracy in measuring the velocity of a particle increases, the accuracy in measuring its position will

When a certain metal is illuminated by light, photoelectrons are observed provided that the wavelength of the light is less than 669 nm. Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10-34 J ∙ s, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J)

A metal having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 4.0 eV. What is the maximum kinetic energy of the photoelectrons produced by this light? (h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J)

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s = 4.141 × 10-15 ev ∙ s, 1 eV = 1.60 × 10-19 J)

If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will

A metal having a work function of 2.8 eV is illuminated with monochromatic light whose photon energy is 3.9 eV. What is the threshold frequency for photoelectron production? (h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J)

A metal surface has a work function of 1.50 eV. Calculate the maximum kinetic energy, in eV, of electrons ejected from this surface by electromagnetic radiation of wavelength 311 nm. (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, e = - 1.60 × 10-19 C, 1 eV = 1.60 × 10-19 J)

Gamma rays are photons with very high energy. How many visible-light photons with a wavelength of 500 nm would you need to match the energy of a gamma-ray photon with energy 4.1 × 10-13 J? (h = 6.626 × 10-34 J ∙ s, c = 3.00 × 108 m/s)

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s)

A beam of red light and a beam of violet light each deliver the same power on a surface. For which beam is the number of photons hitting the surface per second the greatest?

A light beam from a 2.1-mW He-Ne laser has a wavelength of 633 nm. How many photons does the laser emit in one second? (h = 6.626 × 10-34 J ∙ s, c = 3.00 × 108 m/s)

Monochromatic light strikes a metal surface and electrons are ejected from the metal. If the intensity of the light is increased, what will happen to the ejection rate and maximum energy of the electrons?

Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 1.45 eV. What is the work function of this metal? (h = 6.626 × 10-34 J ∙ s, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J)

An 84-kW AM radio station broadcasts at 1000 kHz. How many photons are emitted each second by the transmitting antenna? (h = 6.626 × 10^-34 J ∙ s)

When a certain metal is illuminated by light, photoelectrons are observed provided that the wavelength of the light is less than 669 nm. Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J)

A metal having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 4.0 eV. What is the maximum kinetic energy of the photoelectrons produced by this light? (h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s = 4.141 × 10^-15 ev ∙ s, 1 eV = 1.60 × 10^-19 J)

A metal having a work function of 2.8 eV is illuminated with monochromatic light whose photon energy is 3.9 eV. What is the threshold frequency for photoelectron production? (h = 6.626 × 10^-34J ∙ s, 1 eV = 1.60 × 10^-19J)

A metal surface has a work function of 1.50 eV. Calculate the maximum kinetic energy, in eV, of electrons ejected from this surface by electromagnetic radiation of wavelength 311 nm. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10^-34 J ∙ s, e = - 1.60 × 10^-19 C, 1 eV = 1.60 × 10^-19 J)

Gamma rays are photons with very high energy. How many visible-light photons with a wavelength of 500 nm would you need to match the energy of a gamma-ray photon with energy 4.1 × 10^-13 J? (h = 6.626 × 10^-34J ∙ s, c = 3.00 × 10⁸m/s)

A laser emits light of wavelength 463 nm during a brief pulse that lasts for 25 ms and has a total energy of 1.2 J. How many photons are emitted in that single pulse? (c = 3.00 × 10⁸m/s, h = 6.626 × 10^-34 J ∙ s)

A light beam from a 2.1-mW He-Ne laser has a wavelength of 633 nm. How many photons does the laser emit in one second? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s)

Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 1.45 eV. What is the work function of this metal? (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J)