Question 1

What is the cutoff (threshold)frequency for a metal surface that has a work function of 5.42 eV? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The cutoff frequency is calculated using the formula $ f = \frac{\phi}{h} $, where $ \phi $ is the work function in joules. Converting 5.42 eV to joules gives $ 5.42 \times 1.60 \times 10^{-19} $ J. Dividing by Planck's constant $ h = 6.626 \times 10^{-34} $ J∙s gives approximately $ 1.31 \times 10^{15} $ Hz.

Question 2

What is the longest wavelength of electromagnetic radiation that will eject photoelectrons from sodium metal for which the work function is 2.28 eV? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of What is the longest wavelength of electromagnetic...

Question 3

When it is struck by 240-nm photons,a material having a work function of 2.60 eV emits electrons.What is the maximum kinetic energy of the emitted electrons? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

To find the maximum kinetic energy of the emitted electrons, use the photoelectric effect equation: $KE_{max} = \frac{hc}{\lambda} - \phi$, where $KE_{max}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident photons, and $\phi$ is the work function of the material. First, convert the work function from eV to Joules: $2.60 \, \text{eV} = 2.60 \times 1.60 \times 10^{-19} \, \text{J}$. Then, calculate the energy of the incident photons: $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}{240 \times 10^{-9} \, \text{m}}$. Convert this energy to eV and subtract the work function to find the maximum kinetic energy in eV. The calculation yields approximately 2.58 eV, making choice A correct.

Question 4

What is the longest wavelength of light that can cause photoelectron emission from a metal that has a work function of 2.20 eV? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The wavelength can be found using the equation $ \lambda = \frac{hc}{E} $, where $ h $ is Planck's constant, $ c $ is the speed of light, and $ E $ is the energy in joules. First, convert the work function to joules: $ 2.20 \, \text{eV} = 2.20 \times 1.60 \times 10^{-19} \, \text{J} = 3.52 \times 10^{-19} \, \text{J} $. Then, $ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{3.52 \times 10^{-19}} \approx 5.65 \times 10^{-7} \, \text{m} = 565 \, \text{nm} $.

Question 5

When it is struck by 240-nm photons,a metal ejects electrons with a maximum kinetic energy of $2.58 \mathrm { eV }$ What is the work function of this material? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 16.0 × 10^-19 J)

Accepted Answer

The work function is given by:$$W = hf - KE_{max}$$where:* $h$ is Planck's constant* $f$ is the frequency of the incident photons* $KE_{max}$ is the maximum kinetic energy of the ejected electronsWe can calculate the frequency of the incident photons using the following equation:$$f = \frac{c}{\lambda}$$where:* $c$ is the speed of light* $\lambda$ is the wavelength of the incident photonsSubstituting the given values into the above equations, we get:$$f = \frac{3.00 \times 10^8 \text{ m/s}}{240 \times 10^{-9} \text{ m}} = 1.25 \times 10^{15} \text{ Hz}$$$$W = (6.626 \times 10^{-34} \text{ J s})(1.25 \times 10^{15} \text{ Hz}) - (2.58 \times 16.0 \times 10^{-19} \text{ J})$$$$W = 2.60 \text{ eV}$$Therefore, the work function of the material is 2.60 eV.

Question 6

A photocathode having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 3.4 eV.What is maximum kinetic energy of the photoelectrons produced?(c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

The maximum kinetic energy of the photoelectrons is given by the difference between the photon energy and the work function: $3.4 \, \text{eV} - 2.4 \, \text{eV} = 1.0 \, \text{eV}$. Converting this to joules using $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$, we get $1.0 \times 1.60 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-19} \, \text{J}$.

Question 7

When a metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 520 nm.What is the metal's work function? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

The energy of a photon can be calculated as:

E = hc/λ

where h is Planck's constant, c is the speed of light, and λ is the wavelength of the light.

For photoemission to occur, the energy of the photon must be greater than or equal to the work function (Φ) of the metal, which is the minimum energy required to remove an electron from the metal. This can be expressed as:

E ≥ Φ

Combining these two equations, we get:

hc/λ ≥ Φ

Solving for Φ, we get:

Φ = hc/λ

= (6.626 × 10^-34 J∙s) × (3.00 × 10^8 m/s)/(520 × 10^-9 m)

= 3.81 × 10^-19 J

Converting the work function to electronvolts (eV):

1 eV = 1.60 × 10^-19 J

Φ = 3.81 × 10^-19 J ÷ 1.60 × 10^-19 J/eV

= 2.38 eV

Therefore, the metal's work function is 2.38 eV, which is closest to choice A (2.4 eV).

Question 8

A metal has a work function of 4.50 eV.Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shines on the metal.(1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The energy of a photon of light can be calculated using the formula:

E = hc/λ

where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of light.

We can use this formula to calculate the energy of a photon of light with a wavelength of 250 nm:

E = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (250 × 10^-9 m) = 7.95 × 10^-19 J

To calculate the maximum kinetic energy of the photoelectrons, we can use the formula:

KEmax = E - Φ

where KEmax is the maximum kinetic energy of the photoelectrons, E is the energy of the photon of light, and Φ is the work function of the metal.

Plugging in the values given in the question, we get:

KEmax = (7.95 × 10^-19 J) - (4.50 eV * 1.60 × 10^-19 J/eV) = 0.47 eV

Therefore, the maximum kinetic energy of the photoelectrons is 0.47 eV. The correct choice is C.

Question 9

In her physics laboratory,Mathilda shines electromagnetic radiation on a material and collects photoelectric data to determine Planck's constant.She measures a stopping potential of 5.82 V for radiation of wavelength 100 nm,and 17.99 V for radiation of wavelength 50.0 nm.(1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s) (a)Using Mathilda's data,what value does she determine for Planck's constant? (b)What is the work function of the material Mathilda is using,in electron-volts?

Accepted Answer

The answer of In her physics laboratory,Mathilda shines electromagnetic radiation...

Question 10

When a photoelectric surface is illuminated with light of wavelength 437 nm,the stopping potential is measured to be 1.67 V.(1 eV = 1.60 × 10^-19 J,e = 1.60 × 10^-19 C,m_electron = 9.11 × 10^-31 kg,h = 6.626 × 10^-34 J ∙ s) (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 photoelectric surface is illuminated with...

Question 11

Light with a frequency of 8.70 × 10¹⁴ Hz is incident on a metal that has a work function of 2.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of Light with a frequency of 8.70 ×...

Question 12

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of Light with a wavelength of 310 nm...

Question 13

If the work function of a metal surface is 2.20 eV,what frequency of incident light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface? (1 eV = 1.60 × 10^-19 J, h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of If the work function of a metal...

Question 14

A photocathode whose work function is 2.9 eV is illuminated with white light that has a continuous wavelength band from 400 nm to 700 nm.What is the range of the wavelength band in this white light illumination for which photoelectrons are not produced?(c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of A photocathode whose work function is 2.9...

Question 15

For a certain metal,light of frequency 7.24 × 10^-14 Hz is just barely able to dislodge photoelectrons from the metal.(h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J,e = 1.60 × 10^-19 C) (a)What will be the stopping potential if light of frequency 8.75 × 10^-14 Hz is shone on the metal? (b)What is the work function (in electron-volts)of this metal?

Accepted Answer

The answer of For a certain metal,light of frequency 7.24...

Question 16

The work function of a certain metal is 1.90 eV.What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The answer of The work function of a certain metal...

Question 17

The work function of a particular metal is $4.20 \times 10 ^{- 19} \mathrm {~J}$ What is the photoelectric cutoff (threshold)wavelength for this metal? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

Explanation:The photoelectric cutoff wavelength is given by:$$\lambda = \frac{hc}{\phi}$$where:* h is Planck's constant (6.626 × 10^-34 J ∙ s)* c is the speed of light (3.00 × 10^8 m/s)* ϕ is the work function (4.20 × 10^-19 J)Plugging in the values, we get:$$\lambda = \frac{(6.626 	imes 10^{-34} \mathrm{~J \cdot s})(3.00 	imes 10^8 \mathrm{~m/s})}{4.20 	imes 10^{-19} \mathrm{~J}} = 4.73 	imes 10^{-7} \mathrm{~m} = 473 \mathrm{~nm}$$Therefore, the photoelectric cutoff wavelength for this metal is 473 nm.

Question 18

A photocathode that has a work function of 2.4 eV is illuminated with monochromatic light having photon energy 3.5 eV.What is the wavelength of this light? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The energy of the incident light is 3.5 eV. Since 1 eV = 1.60 × 10-19 J, the energy of the photon is:

E = 3.5 eV × 1.60 × 10-19 J/eV = 5.60 × 10-19 J

The work function of the photocathode is 2.4 eV, so the minimum energy required to emit an electron is:

ϕ = 2.4 eV × 1.60 × 10-19 J/eV = 3.84 × 10-19 J

The difference between the energy of the photon and the work function is the maximum kinetic energy of the emitted electron:

K_emax = E - ϕ = 5.60 × 10-19 J - 3.84 × 10-19 J = 1.76 × 10-19 J

The momentum of the electron is related to its kinetic energy by:

K_emax = p2/(2m)

where p is the momentum and m is the mass of the electron. Since the momentum is related to the wavelength of the photon by:

p = h/λ

we can combine these equations to get:

λ = h/(2mK_emax) = (6.626 × 10-34 J∙s)/(2 × 9.11 × 10-31 kg × 1.76 × 10-19 J) = 3.50 × 10-7 m = 350 nm

Therefore, the wavelength of the incident light is 350 nm. The correct answer is A.

Question 19

A photocathode having a work function of 2.8 eV is illuminated with monochromatic electromagnetic radiation whose photon energy is 4.0 eV.What is the threshold (cutoff)frequency for photoelectron production? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Accepted Answer

The threshold frequency is given by the equation:$$f_0 = \frac{\phi}{h}$$where:* $f_0$ is the threshold frequency* $\phi$ is the work function* $h$ is Planck's constantSubstituting the given values into the equation, we get:$$f_0 = \frac{2.8 	ext{ eV}}{6.626 	imes 10^{-34} 	ext{ J s}} = 6.8 	imes 10^{14} 	ext{ Hz}$$Therefore, the correct answer is A.

Question 20

A beam of light with a frequency range from 3.01 × 10¹⁴ Hz to 6.10 × 10¹⁴ Hz is incident on a metal surface.If the work function of the metal surface is 2.20 eV,what is the maximum kinetic energy of photoelectrons ejected from this surface? (h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

Accepted Answer

The answer of A beam of light with a frequency...

Quiz 27: Quantum Optics

What is the cutoff (threshold) frequency for a metal surface that has a work function of 5.42 eV? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

What is the longest wavelength of electromagnetic radiation that will eject photoelectrons from sodium metal for which the work function is 2.28 eV? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

When it is struck by 240-nm photons,a material having a work function of 2.60 eV emits electrons.What is the maximum kinetic energy of the emitted electrons? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

What is the longest wavelength of light that can cause photoelectron emission from a metal that has a work function of 2.20 eV? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

When it is struck by 240-nm photons,a metal ejects electrons with a maximum kinetic energy of $2.58 \mathrm { eV }$ What is the work function of this material? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 16.0 × 10^-19 J)

A photocathode having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 3.4 eV.What is maximum kinetic energy of the photoelectrons produced?(c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

When a metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 520 nm.What is the metal's work function? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

A metal has a work function of 4.50 eV.Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shines on the metal.(1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Light with a frequency of 8.70 × 10¹⁴ Hz is incident on a metal that has a work function of 2.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

If the work function of a metal surface is 2.20 eV,what frequency of incident light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface? (1 eV = 1.60 × 10^-19 J, h = 6.626 × 10^-34 J ∙ s)

The work function of a certain metal is 1.90 eV.What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

The work function of a particular metal is $4.20 \times 10 ^{- 19} \mathrm {~J}$ What is the photoelectric cutoff (threshold) wavelength for this metal? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

A photocathode that has a work function of 2.4 eV is illuminated with monochromatic light having photon energy 3.5 eV.What is the wavelength of this light? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

A photocathode having a work function of 2.8 eV is illuminated with monochromatic electromagnetic radiation whose photon energy is 4.0 eV.What is the threshold (cutoff) frequency for photoelectron production? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Quiz 27: Quantum Optics

What is the cutoff (threshold) frequency for a metal surface that has a work function of 5.42 eV? (1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

What is the longest wavelength of electromagnetic radiation that will eject photoelectrons from sodium metal for which the work function is 2.28 eV? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J)

When it is struck by 240-nm photons,a material having a work function of 2.60 eV emits electrons.What is the maximum kinetic energy of the emitted electrons? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

What is the longest wavelength of light that can cause photoelectron emission from a metal that has a work function of 2.20 eV? (1 eV = 1.60 × 10-19 J,c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)

When it is struck by 240-nm photons,a metal ejects electrons with a maximum kinetic energy of 2.58eV2.58 \mathrm { eV }2.58eV What is the work function of this material? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 16.0 × 10-19 J)

A photocathode having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 3.4 eV.What is maximum kinetic energy of the photoelectrons produced?(c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

When a metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 520 nm.What is the metal's work function? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

A metal has a work function of 4.50 eV.Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shines on the metal.(1 eV = 1.60 × 10-19 J,c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)

Light with a frequency of 8.70 × 1014 Hz is incident on a metal that has a work function of 2.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10-19 J,c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)

If the work function of a metal surface is 2.20 eV,what frequency of incident light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface? (1 eV = 1.60 × 10-19 J, h = 6.626 × 10-34 J ∙ s)

The work function of a certain metal is 1.90 eV.What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10-19 J,c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)

The work function of a particular metal is 4.20×10−19 J4.20 \times 10 ^{- 19} \mathrm {~J}4.20×10−19 J What is the photoelectric cutoff (threshold) wavelength for this metal? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)

A photocathode that has a work function of 2.4 eV is illuminated with monochromatic light having photon energy 3.5 eV.What is the wavelength of this light? (c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J)

A photocathode having a work function of 2.8 eV is illuminated with monochromatic electromagnetic radiation whose photon energy is 4.0 eV.What is the threshold (cutoff) frequency for photoelectron production? (1 eV = 1.60 × 10-19 J,h = 6.626 × 10-34 J ∙ s)

A beam of light with a frequency range from 3.01 × 1014 Hz to 6.10 × 1014 Hz is incident on a metal surface.If the work function of the metal surface is 2.20 eV,what is the maximum kinetic energy of photoelectrons ejected from this surface? (h = 6.626 × 10-34 J ∙ s,1 eV = 1.60 × 10-19 J)

What is the cutoff (threshold) frequency for a metal surface that has a work function of 5.42 eV? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

What is the longest wavelength of electromagnetic radiation that will eject photoelectrons from sodium metal for which the work function is 2.28 eV? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

When it is struck by 240-nm photons,a material having a work function of 2.60 eV emits electrons.What is the maximum kinetic energy of the emitted electrons? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

What is the longest wavelength of light that can cause photoelectron emission from a metal that has a work function of 2.20 eV? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

When it is struck by 240-nm photons,a metal ejects electrons with a maximum kinetic energy of $2.58 \mathrm { eV }$ What is the work function of this material? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 16.0 × 10^-19 J)

A photocathode having a work function of 2.4 eV is illuminated with monochromatic light whose photon energy is 3.4 eV.What is maximum kinetic energy of the photoelectrons produced?(c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

When a metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 520 nm.What is the metal's work function? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19 J)

A metal has a work function of 4.50 eV.Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shines on the metal.(1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

Light with a frequency of 8.70 × 10¹⁴ Hz is incident on a metal that has a work function of 2.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)

Light with a wavelength of 310 nm is incident on a metal that has a work function of 3.80 eV.What is the maximum kinetic energy that a photoelectron ejected in this process can have? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

If the work function of a metal surface is 2.20 eV,what frequency of incident light would give a maximum kinetic energy of 0.25 eV to the photoelectrons ejected from this surface? (1 eV = 1.60 × 10^-19 J, h = 6.626 × 10^-34 J ∙ s)

The work function of a certain metal is 1.90 eV.What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10^-19 J,c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

The work function of a particular metal is $4.20 \times 10 ^{- 19} \mathrm {~J}$ What is the photoelectric cutoff (threshold) wavelength for this metal? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s)

A photocathode that has a work function of 2.4 eV is illuminated with monochromatic light having photon energy 3.5 eV.What is the wavelength of this light? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s, 1 eV = 1.60 × 10^-19 J)

A photocathode having a work function of 2.8 eV is illuminated with monochromatic electromagnetic radiation whose photon energy is 4.0 eV.What is the threshold (cutoff) frequency for photoelectron production? (1 eV = 1.60 × 10^-19 J,h = 6.626 × 10^-34 J ∙ s)