Question 1

Of the following photons,which has the lowest energy?
A)visible
B)ultraviolet
C)infrared
D)microwave

Accepted Answer

The energy of a photon is inversely proportional to its wavelength. Therefore, of the given options, the photon with the longest wavelength will have the lowest energy. A microwave photon has the longest wavelength, hence it has the lowest energy.

Question 2

What is the energy of a photon whose frequency is 2.9 × 10²⁰ Hz? (h = 6.63 × 10−³⁴ J⋅s and 1 eV = 1.60 × 10−¹⁹ J) A)0.2 MeV B)1.2 MeV C)1.9 MeV D)3.1 MeV E)0.8 MeV

Accepted Answer

The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency. Substituting the given values, E = (6.63 × 10^-34 J⋅s)(2.9 × 10^20 Hz) = 1.9227 × 10^-13 J. Converting this to MeV using 1 eV = 1.60 × 10^-19 J, we get E ≈ 1.2 MeV.

Question 3

Which change will not change the kinetic energy of the most energetic electrons emitted in the photoelectric effect?
A)changing the brightness of the light
B)changing the frequency of the light
C)changing the metal the light is hitting
D)All of the above will change the electron's kinetic energy.

Accepted Answer

The kinetic energy of the most energetic electrons in the photoelectric effect is determined by the frequency of the incident light and the work function of the metal, not by the brightness of the light. Changing the brightness only affects the number of electrons emitted, not their kinetic energy.

Question 4

How much energy (in eV)does a photon of orange light (λ = 609 nm)have? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)3.27 eV B)0.31 eV C)2.04 eV D)0.68 eV E)5.23 eV

Accepted Answer

First, we need to calculate the energy of the photon using the equation:

E = hc/λ

where:
h = Planck's constant = 6.63 × 10^-34 J·s
c = speed of light = 3.00 × 10^8 m/s
λ = wavelength of orange light = 609 nm = 609 × 10^-9 m

Substituting the values, we get:

E = (6.63 × 10^-34 J·s × 3.00 × 10^8 m/s) / (609 × 10^-9 m)
E = 3.42 × 10^-19 J

Now, we need to convert this energy to electron volts (eV) using the conversion factor:

1 eV = 1.60 × 10^-19 J

So, dividing the energy by the conversion factor, we get:

E = (3.42 × 10^-19 J) / (1.60 × 10^-19 J/eV)
E = 2.14 eV

Therefore, the answer is C) 2.04 eV (rounding off to two significant figures).

Question 5

Of the following energies for photons,which is the least energy that could result in photoelectron production if the work function is 3.77 eV?
A)2.72 eV
B)3.19 eV
C)4.57 eV
D)6.98 eV
E)​6.53 eV

Accepted Answer

The energy required to produce a photoelectron must be equal to or greater than the work function. Among the given options, 4.57 eV is the least energy that is greater than the work function of 3.77 eV, thus capable of resulting in photoelectron production.

Question 6

What is the wavelength of a monochromatic light beam,where the photon energy is 4.14 × 10−¹⁹ J? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m) A)160 nm B)480 nm C)72 nm D)187 nm E)915 nm

Accepted Answer

The answer of What is the wavelength of a monochromatic...

Question 7

Violet light (λ = 430 nm)is incident on a piece of potassium (φ = 2.20 eV).What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)2.43 eV B)0.69 eV C)2.89 eV D)5.09 eV E)0.96 eV

Accepted Answer

The maximum kinetic energy of the ejected photoelectrons is given by:
KEmax = hν - φ
We know that:
ν = c/λ
Therefore:
ν = (3.00 × 108 m/s)/(430 × 10-9 m)
ν = 6.98 × 1014 Hz
We can substitute this value, along with the given values of h and φ, into the above equation to get:
KEmax = (6.63 × 10-34 J s)(6.98 × 1014 Hz) - (2.20 eV)(1.60 × 10-19 J/eV)
KEmax = 4.62 × 10-19 J - 3.52 × 10-19 J
KEmax = 1.10 × 10-19 J
We can convert this value to electronvolts (eV) by dividing by the conversion factor 1.60 × 10-19 J/eV:
KEmax = (1.10 × 10-19 J)/(1.60 × 10-19 J/eV)
KEmax = 0.69 eV

Therefore, the correct answer is B.

Question 8

Light of wavelength 439 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 1.2 V.What is the work function of the metal? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)1.63 eV B)2.83 eV C)4.03 eV D)0.82 eV E)2.02 eV

Accepted Answer

The work function ($\phi$) of the metal can be calculated using the photoelectric equation: $\phi = E - eV$, where $E$ is the energy of the incident photon, $eV$ is the energy corresponding to the stopping potential, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light. First, calculate the energy of the incident photon: $E = \frac{hc}{\lambda}$. Substituting the given values, $E = \frac{6.63 \times 10^{-34} \text{ J⋅s} \times 3.00 \times 10^8 \text{ m/s}}{439 \times 10^{-9} \text{ m}}$. This gives $E \approx 4.53 \times 10^{-19} \text{ J}$. Converting this energy to electron volts (eV) by dividing by the charge of an electron (1.60 \times 10^{-19} J/eV) gives $E \approx 2.83 \text{ eV}$. The energy corresponding to the stopping potential is $eV = 1.2 \text{ eV}$ (since 1 eV = 1.60 \times 10^{-19} J and the stopping potential is 1.2 V). Therefore, the work function is $\phi = E - eV = 2.83 \text{ eV} - 1.2 \text{ eV} = 1.63 \text{ eV}$.

Question 9

Who was the first to successfully explain the photoelectric effect?
A)Planck
B)Young
C)Bohr
D)Einstein

Accepted Answer

Albert Einstein was the first to successfully explain the photoelectric effect in 1905, which earned him the Nobel Prize in Physics in 1921.

Question 10

A sodium vapor lamp has a power output of 60 W.If 547 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m) A)10¹⁸ B)10²⁰ C)10³⁹ D)10²⁹ E)10²⁸

Accepted Answer

First, calculate the energy of one photon using E = hc/λ. Then, divide the total power output by the energy per photon to find the number of photons emitted per second.

Question 11

A helium-neon laser emits red light having a wavelength of 632.8 nm and a power of 1.85 mW.How many photons are emitted each second? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m) A)5.9E+15 B)3.7E+34 C)5.9E+24 D)5.9E+18 E)1.2E+17

Accepted Answer

To find the number of photons emitted per second, we first convert the wavelength to meters: 632.8 nm = 632.8 × 10^-9 m. The energy of a single photon is given by E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J⋅s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength in meters. Substituting the given values, we get E = (6.63 × 10^-34 J⋅s)(3.00 × 10^8 m/s) / (632.8 × 10^-9 m) = 3.14 × 10^-19 J per photon. The power of the laser is 1.85 mW, which is 1.85 × 10^-3 J/s. The number of photons emitted per second is the total energy per second divided by the energy per photon: (1.85 × 10^-3 J/s) / (3.14 × 10^-19 J) = 5.9 × 10^15 photons per second.

Question 12

What is the wavelength of a monochromatic light beam,where the photon energy is 2.15 eV? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 nm = 10−⁹ m,and 1 eV = 1.6 × 10−¹⁹ J) A)308 nm B)578 nm C)223 nm D)156 nm E)475 nm

Accepted Answer

The answer of What is the wavelength of a monochromatic...

Question 13

A photon absorbed by an electron will give up more energy to the electron if the photon:
A)is not spread out over many electrons.
B)is moving faster.
C)is moving slower.
D)has a higher frequency.

Accepted Answer

The energy of a photon is directly proportional to its frequency. Therefore, a photon with a higher frequency (and therefore higher energy) will give up more energy to an electron upon absorption.

Question 14

Of photons of red,yellow,light,and blue light,which photons have the greatest energy?
A)red
B)yellow
C)green
D)blue

Accepted Answer

The energy of a photon is directly proportional to its frequency, and blue light has the highest frequency (and therefore the highest energy) among the given options, followed by green, yellow, and red in decreasing order of frequency and energy.

Question 15

Sources of red,blue,and yellow light each emit light with a power of 50 mW.Which source emits more photons per second?
A)the red source
B)the blue source
C)the yellow source
D)They all emit the same number per second.

Accepted Answer

The energy of a photon is inversely proportional to its wavelength. Red light has the longest wavelength among the three, meaning it has the lowest energy per photon. Therefore, for the same power, the red source emits more photons per second.

Question 16

What is the frequency of monochromatic light where the photon energy is 4.3 × 10−¹⁹ J? (h = 6.63 × 10−³⁴ J⋅s) A)4.3E+15 Hz B)2.1E+15 Hz C)6.5E+14 Hz D)1.9E+14 Hz E)6.5E+13 Hz

Accepted Answer

The frequency of monochromatic light ($f$) can be calculated using the formula for the energy of a photon: $E = hf$, where $E$ is the energy of the photon, $h$ is Planck's constant, and $f$ is the frequency of the light. Rearranging the formula to solve for frequency gives $f = \frac{E}{h}$. Substituting the given values, $f = \frac{4.3 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.5 \times 10^{14} \, \text{Hz}$.

Question 17

A light beam is shining on a metal target that has a work function of 1.8 eV.If a stopping potential of 0.7 V is required,what is the wavelength of the incoming monochromatic light? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)497 nm B)166 nm C)796 nm D)1130 nm E)691 nm

Accepted Answer

The energy of the incoming light can be calculated using the photoelectric effect equation: $E = \phi + eV$, where $E$ is the energy of the incoming light, $\phi$ is the work function, and $V$ is the stopping potential. Substituting the given values, $E = 1.8\, \text{eV} + 0.7\, \text{eV} = 2.5\, \text{eV}$. Converting this energy to joules gives $E = 2.5 \times 1.60 \times 10^{-19}\, \text{J} = 4.0 \times 10^{-19}\, \text{J}$. The wavelength $\lambda$ can be found using the equation $E = \frac{hc}{\lambda}$, solving for $\lambda$ gives $\lambda = \frac{hc}{E}$. Substituting $h = 6.63 \times 10^{-34}\, \text{J⋅s}$, $c = 3.00 \times 10^8\, \text{m/s}$, and $E = 4.0 \times 10^{-19}\, \text{J}$ gives $\lambda \approx 497\, \text{nm}$.

Question 18

An ultraviolet light beam having a wavelength of 217 nm is incident on a molybdenum surface with work function of 4.9 eV.What is the stopping potential? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.6 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)10.6 V B)5.3 V C)0.8 V D)0.4 V E)5.7 V

Accepted Answer

The answer of An ultraviolet light beam having a wavelength...

Question 19

Light of wavelength 6.9 × 10−⁷ m has an energy of: (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s) A)2.9E-19 J B)9.6E-20 J C)3.1E-19 J D)4.3E-20 J E)3.2E-20 J

Accepted Answer

The energy of a photon is calculated using E = hc/λ. Substituting the given values: E = (6.63 × 10^-34 J⋅s)(3.00 × 10^8 m/s) / (6.9 × 10^-7 m) = 2.9 × 10^-19 J.

Question 20

Of photons of red,yellow,green,and blue light,which photons have the least energy?
A)red
B)blue
C)green
D)yellow

Accepted Answer

The energy of a photon is directly proportional to its frequency, and inversely proportional to its wavelength. Red light has the longest wavelength and lowest frequency among the given options, therefore it has the least energy.

Quiz 27: Quantum Physics

Of the following photons,which has the lowest energy?

What is the energy of a photon whose frequency is 2.9 × 10²⁰ Hz? (h = 6.63 × 10−³⁴ J⋅s and 1 eV = 1.60 × 10−¹⁹ J)

Which change will not change the kinetic energy of the most energetic electrons emitted in the photoelectric effect?

How much energy (in eV) does a photon of orange light (λ = 609 nm) have? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Of the following energies for photons,which is the least energy that could result in photoelectron production if the work function is 3.77 eV?

What is the wavelength of a monochromatic light beam,where the photon energy is 4.14 × 10−¹⁹ J? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m)

Violet light (λ = 430 nm) is incident on a piece of potassium (φ = 2.20 eV) .What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Light of wavelength 439 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 1.2 V.What is the work function of the metal? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Who was the first to successfully explain the photoelectric effect?

A sodium vapor lamp has a power output of 60 W.If 547 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m)

A helium-neon laser emits red light having a wavelength of 632.8 nm and a power of 1.85 mW.How many photons are emitted each second? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,and 1 nm = 10−⁹ m)

What is the wavelength of a monochromatic light beam,where the photon energy is 2.15 eV? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 nm = 10−⁹ m,and 1 eV = 1.6 × 10−¹⁹ J)

A photon absorbed by an electron will give up more energy to the electron if the photon:

Of photons of red,yellow,light,and blue light,which photons have the greatest energy?

Sources of red,blue,and yellow light each emit light with a power of 50 mW.Which source emits more photons per second?

What is the frequency of monochromatic light where the photon energy is 4.3 × 10−¹⁹ J? (h = 6.63 × 10−³⁴ J⋅s)

A light beam is shining on a metal target that has a work function of 1.8 eV.If a stopping potential of 0.7 V is required,what is the wavelength of the incoming monochromatic light? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

An ultraviolet light beam having a wavelength of 217 nm is incident on a molybdenum surface with work function of 4.9 eV.What is the stopping potential? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.6 × 10−¹⁹ J,and 1 nm = 10−⁹ m)

Light of wavelength 6.9 × 10−⁷ m has an energy of: (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s)

Of photons of red,yellow,green,and blue light,which photons have the least energy?

Quiz 27: Quantum Physics

Of the following photons,which has the lowest energy?

What is the energy of a photon whose frequency is 2.9 × 1020 Hz? (h = 6.63 × 10−34 J⋅s and 1 eV = 1.60 × 10−19 J)

Which change will not change the kinetic energy of the most energetic electrons emitted in the photoelectric effect?

How much energy (in eV) does a photon of orange light (λ = 609 nm) have? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,1 eV = 1.60 × 10−19 J,and 1 nm = 10−9 m)

Of the following energies for photons,which is the least energy that could result in photoelectron production if the work function is 3.77 eV?

What is the wavelength of a monochromatic light beam,where the photon energy is 4.14 × 10−19 J? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,and 1 nm = 10−9 m)

Violet light (λ = 430 nm) is incident on a piece of potassium (φ = 2.20 eV) .What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,1 eV = 1.60 × 10−19 J,and 1 nm = 10−9 m)

Light of wavelength 439 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 1.2 V.What is the work function of the metal? (h = 6.63 × 10−34 J⋅s,c = 3.00 × 108 m/s,1 eV = 1.60 × 10−19 J,and 1 nm = 10−9 m)