Question 1

What is the minimum x-ray wavelength produced when electrons are accelerated through a potential of 50 000 V? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,and 1 eV = 1.60 x 10^-19J) A)12.4 x 10^-12 m B)16.5 x 10^-12 m C)19.8 x 10^-12 m D)24.9 x 10^-12 m

Accepted Answer

The answer of What is the minimum x-ray wavelength produced...

Question 2

An ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with work function of 4.2 eV.What is the stopping potential? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.6 x 10^-19 J,and 1 nm = 10^-9 m) A)1.3 V B)3.5 V C)5.4 V D)11.9 V

Accepted Answer

The energy of a photon can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ (lambda) is the wavelength of the photon. Plugging in the given values, we get:

E = (6.63 x 10^-34 J*s)(3.00 x 10^8 m/s)/(130 x 10^-9 m)
E = 1.52 x 10^-18 J

The stopping potential can be calculated using the equation V_stop = (E_photon - W)/e, where E_photon is the energy of the photon, W is the work function, and e is the elementary charge. Plugging in the given values, we get:

V_stop = (1.52 x 10^-18 J - (4.2 eV)(1.6 x 10^-19 J/e))/(1.6 x 10^-19 C/e)
V_stop ≈ 5.4 V

Therefore, the best choice is C, with a stopping potential of 5.4 V.

Question 3

Blue light (l = 460 nm)is incident on a piece of potassium (f = 2.20 eV).What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9 m) A)1.0 eV B)0.50 eV C)0.25 eV D)4.9 eV

Accepted Answer

The answer of Blue light (l = 460 nm)is incident...

Question 4

Which of the following statements best describes the relation between the quantum theory and the photoelectric effect experiment?
A)Quantum theory explains the photoelectric effect.
B)The photoelectric effect contradicts quantum theory.
C)Quantum theory has no bearing on the photoelectric effect.
D)The photoelectric effect explains quantum theory.

Accepted Answer

The photoelectric effect experiment provided crucial evidence supporting the quantum theory. The theory proposed that electromagnetic radiation (light) was made up of particles, photons, with discrete energy levels. The photoelectric effect experiment, which involved shining light on a metal surface and measuring the resulting electron emission, demonstrated that the energy of the emitted electrons was directly proportional to the frequency of the incoming photons. This relationship could only be explained by the quantum theory, which predicted that the energy of a photon would be proportional to its frequency. Therefore, quantum theory explains the photoelectric effect.

Question 5

A light beam is shining on a metal target that has a work function of 2.20 eV.If a stopping potential of 1.30 V is required,what is the wavelength of the incoming monochromatic light? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9 m) A)355 nm B)497 nm C)744 nm D)1 421 nm

Accepted Answer

The energy of the incoming light is the sum of the work function and the stopping potential energy. Convert the work function and stopping potential to joules, then use E = hf = hc/λ to find the wavelength.

Question 6

Light of wavelength 450 nm is incident on a target metal that has a work function of 1.80 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9 m) A)0.57 V B)0.96 V C)2.76 V D)4.56 V

Accepted Answer

The stopping potential can be calculated by:
$V_s = \frac{hc}{\lambda e} - \phi$
where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of light, $e$ is the charge of an electron, and $\phi$ is the work function of the metal.

Plugging in the given values, we get:
$V_s = \frac{(6.63 	imes 10^{-34} J \cdot s)(3.00 	imes 10^8 m/s)}{(450 	imes 10^{-9} m)(1.60 	imes 10^{-19} C)} - 1.80 eV$

Simplifying this, we get:
$V_s = 0.96 V$

Therefore, the best choice is (B).

Question 7

How much energy (in eV)does a photon of red light (l = 700 nm)have? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9m) A)3.11 eV B)2.26 eV C)1.78 eV D)1.24 eV

Accepted Answer

Firstly, we need to convert the wavelength of red light from nanometers to meters:

λ = 700 nm = 700 x 10^-9 m

Next, we can use the equation E = hc/λ to calculate the energy of the photon:

E = hc/λ = (6.63 x 10^-34 J s) x (3.00 x 10^8 m/s) / (700 x 10^-9 m)

E = 2.84 x 10^-19 J

Finally, we can convert this energy from joules to electron volts (eV), using the conversion factor 1 eV = 1.60 x 10^-19 J:

E = 2.84 x 10^-19 J x (1 eV/1.60 x 10^-19 J)

E = 1.78 eV

Therefore, the correct answer is C.

Question 8

A helium-neon laser emits red light having a wavelength of 632.8 nm and a power of 0.50 mW.How many photons are emitted each second? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸m/s,and 1 nm = 10^-9m) A)1.6 x 10¹⁵ B)3.3 x 10¹⁶ C)4.8 x 10¹⁷ D)2.6 x 10¹⁸

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 power of the laser is given by P = N E, where N is the number of photons emitted per second. Substituting the given values, we get:P = N E0.50 mW = N (6.63 x 10^-34 J s) (3.00 x 10^8 m/s) / (632.8 x 10^-9 m)N = 1.6 x 10^15 photons/s

Question 9

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

Accepted Answer

Einstein was the first to successfully explain the photoelectric effect in 1905, for which he was awarded the Nobel Prize in Physics in 1921.

Question 10

What is the energy of a photon whose frequency is 6.0 x 10²⁰ Hz? (h = 6.63 x 10^-34 J*s and 1 eV = 1.60 x 10^-19J) A)1.6 MeV B)2.5 MeV C)3.3 MeV D)4.8 MeV

Accepted Answer

The answer of What is the energy of a photon...

Question 11

If barium has a work function of 2.60 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J and 1 nm = 10^-9 m) A)398 nm B)478 nm C)497 nm D)596 nm

Accepted Answer

The cutoff wavelength is calculated using the formula λ = hc/Φ, where Φ is the work function. Converting the work function from eV to Joules (2.60 eV = 2.60 x 1.60 x 10^-19 J), and using h = 6.63 x 10^-34 J·s and c = 3.00 x 10^8 m/s, we find λ = (6.63 x 10^-34 J·s * 3.00 x 10^8 m/s) / (2.60 x 1.60 x 10^-19 J) = 478 nm.

Question 12

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 13

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 answer of Sources of red,blue,and yellow light each emit...

Question 14

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 according to the formula E=hf, where E is the energy, h is Planck's constant, and f is the frequency. Therefore, a photon with a higher frequency will give up more energy to an electron upon absorption.

Question 15

Of the following energies for photons,which is the least energy that could result in photoelectron production if the work function is 3.00 eV?
A)1.50 eV
B)2.90 eV
C)3.50 eV
D)6.01 eV

Accepted Answer

The energy of the photon must be equal to or greater than the work function (3.00 eV) to produce photoelectrons. 3.50 eV is the smallest energy greater than 3.00 eV.

Question 16

Light of wavelength 480 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 0.55 V.What is the work function of the metal? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9m) A)2.04 eV B)3.19 eV C)2.59 eV D)0.55 eV

Accepted Answer

Explanation:The work function of the metal is given by:$$W = h
u - eVs$$where:* h is Planck's constant* $
u$ is the frequency of the light* e is the charge of an electron* Vs is the stopping potentialSubstituting the given values, we get:$$W = (6.63 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m/s}) / (480 	imes 10^{-9} 	ext{ m}) - (1.60 	imes 10^{-19} 	ext{ J})(0.55 	ext{ V})$$$$W = 4.14 	imes 10^{-19} 	ext{ J} - 8.80 	imes 10^{-20} 	ext{ J}$$$$W = 3.26 	imes 10^{-19} 	ext{ J}$$Converting to electron volts, we get:$$W = (3.26 	imes 10^{-19} 	ext{ J}) / (1.60 	imes 10^{-19} 	ext{ J/eV})$$$$W = 2.04 	ext{ eV}$$Therefore, the work function of the metal is 2.04 eV.

Question 17

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

Accepted Answer

Ultraviolet photons have a higher energy than visible or infrared photons, and microwave photons have the lowest energy.

Question 18

According to Einstein,increasing the brightness of a beam of light without changing its color will increase:
A)the number of photons.
B)the energy of each photon.
C)the speed of the photons.
D)the frequency of the photons.

Accepted Answer

Increasing the brightness of a beam of light, according to Einstein's photoelectric effect theory, increases the number of photons in the beam without affecting the energy of each photon, the speed of the photons (which is constant at the speed of light in vacuum), or the frequency of the photons (which is determined by the color or wavelength of the light).

Question 19

What is the maximum velocity of a photoelectron emitted from a surface with work function 5.00 eV when illuminated by 200 nm ultraviolet light? (m_electron = 9.11 x 10^-31 kg,h = 6.63 x 10^-34 J*s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9m) A)800 000 m/s B)653 000 m/s C)431 000 m/s D)212 000 m/s

Accepted Answer

The formula for maximum velocity of a photoelectron is given by $v_{max}=\sqrt{\dfrac{2(E-E_F)}{m}}$ where $E$ is the energy of incident photon, $E_F$ is the work function, and $m$ is the mass of electron.

Converting the given wavelength to energy: $E=\dfrac{hc}{\lambda}=\dfrac{(6.63	imes10^{-34}	ext{ Js})(3	imes10^8	ext{ m/s})}{(200	imes10^{-9}	ext{ m})}=9.94	imes10^{-19}	ext{ J}=6.21	ext{ eV}$

Substituting the given values: 
$v_{max}=\sqrt{\dfrac{2(E-E_F)}{m}}=\sqrt{\dfrac{2(6.21-5.00)	ext{ eV}}{9.11	imes10^{-31}	ext{ kg}}}=653,032	ext{ m/s}\approx \boxed{	extbf{(B)}}$

Question 20

A sodium vapor lamp has a power output of 300 W.If 590 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,and 1 nm = 10^-9 m) A)10¹⁷ B)10²¹ C)10²⁵ D)10²⁹

Accepted Answer

First, we need to find the energy of a single photon with the given wavelength.
Using the formula:
E = hc/λ
where h is Planck's constant, c is the speed of light, and λ is the wavelength
E = (6.63 x 10^-34 J s)(3.00 x 10^8 m/s)/(590 x 10^-9 m)
E = 3.36 x 10^-19 J
Next, we can find the number of photons emitted per second using the power of the lamp and the energy per photon.
Using the formula:
Power = Energy per photon x Number of photons per second
Number of photons per second = Power/Energy per photon
Number of photons per second = (300 W)/(3.36 x 10^-19 J)
Number of photons per second = 8.93 x 10^20
Rounding to three significant figures gives 1.21 x 10^21 or 10^21 photons per second, which corresponds to choice B.

Quiz 27: Quantum Physics

What is the minimum x-ray wavelength produced when electrons are accelerated through a potential of 50 000 V? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,and 1 eV = 1.60 x 10^-19J)

An ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with work function of 4.2 eV.What is the stopping potential? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.6 x 10^-19 J,and 1 nm = 10^-9 m)

Blue light (l = 460 nm) is incident on a piece of potassium (f = 2.20 eV) .What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9 m)

Which of the following statements best describes the relation between the quantum theory and the photoelectric effect experiment?

A light beam is shining on a metal target that has a work function of 2.20 eV.If a stopping potential of 1.30 V is required,what is the wavelength of the incoming monochromatic light? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9 m)

Light of wavelength 450 nm is incident on a target metal that has a work function of 1.80 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9 m)

How much energy (in eV) does a photon of red light (l = 700 nm) have? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9m)

A helium-neon laser emits red light having a wavelength of 632.8 nm and a power of 0.50 mW.How many photons are emitted each second? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸m/s,and 1 nm = 10^-9m)

Who was the first to successfully explain the photoelectric effect?

What is the energy of a photon whose frequency is 6.0 x 10²⁰ Hz? (h = 6.63 x 10^-34 J*s and 1 eV = 1.60 x 10^-19J)

If barium has a work function of 2.60 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J and 1 nm = 10^-9 m)

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

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

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

Of the following energies for photons,which is the least energy that could result in photoelectron production if the work function is 3.00 eV?

Light of wavelength 480 nm is incident on a metallic surface with a resultant photoelectric stopping potential of 0.55 V.What is the work function of the metal? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J,and 1 nm = 10^-9m)

Of the following photons,which has the highest energy?

According to Einstein,increasing the brightness of a beam of light without changing its color will increase:

What is the maximum velocity of a photoelectron emitted from a surface with work function 5.00 eV when illuminated by 200 nm ultraviolet light? (m_electron = 9.11 x 10^-31 kg,h = 6.63 x 10^-34 J*s,1 eV = 1.60 x 10^-19J,and 1 nm = 10^-9m)

A sodium vapor lamp has a power output of 300 W.If 590 nm is the average wavelength of the source,about how many photons are emitted per second? (h = 6.63 x 10^-34 J*s,c = 3.00 x 10⁸ m/s,and 1 nm = 10^-9 m)

Quiz 27: Quantum Physics

What is the minimum x-ray wavelength produced when electrons are accelerated through a potential of 50 000 V? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,and 1 eV = 1.60 x 10-19 J)

An ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with work function of 4.2 eV.What is the stopping potential? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,1 eV = 1.6 x 10-19 J,and 1 nm = 10-9 m)

Blue light (l = 460 nm) is incident on a piece of potassium (f = 2.20 eV) .What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,1 eV = 1.60 x 10-19 J,and 1 nm = 10-9 m)

Which of the following statements best describes the relation between the quantum theory and the photoelectric effect experiment?

A light beam is shining on a metal target that has a work function of 2.20 eV.If a stopping potential of 1.30 V is required,what is the wavelength of the incoming monochromatic light? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,1 eV = 1.60 x 10-19 J,and 1 nm = 10-9 m)

Light of wavelength 450 nm is incident on a target metal that has a work function of 1.80 eV.What stopping potential is required for this combination in a phototube? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,1 eV = 1.60 x 10-19 J,and 1 nm = 10-9 m)

How much energy (in eV) does a photon of red light (l = 700 nm) have? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,1 eV = 1.60 x 10-19 J,and 1 nm = 10-9 m)

A helium-neon laser emits red light having a wavelength of 632.8 nm and a power of 0.50 mW.How many photons are emitted each second? (h = 6.63 x 10-34 J*s,c = 3.00 x 108 m/s,and 1 nm = 10-9 m)