Question 1

An electron in the hydrogen atom makes a transition from an energy level of -1.51 eV to one of energy -3.40 eV.In this process a photon is emitted with what wavelength?
A)1.89 nm
B)656 nm
C)456 nm
D)821 nm
E)365 nm

Accepted Answer

We can use the formula for the energy of a photon: 
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 the photon. 
The energy difference between the two energy levels of the electron is: 
ΔE = -3.40 eV - (-1.51 eV) = -1.89 eV 
We need to convert this energy difference to joules, so we can use the formula: 
1 eV = 1.602 x 10^-19 J 
ΔE = (-1.89 eV) x (1.602 x 10^-19 J/eV) = -3.03 x 10^-19 J 
Now we can plug this into the formula for the energy of the photon: 
-3.03 x 10^-19 J = (6.626 x 10^-34 J.s)(3.00 x 10^8 m/s)/λ 
Solving for λ gives: 
λ = 656 nm

Question 2

A gamma-ray photon of energy 800 keV scatters off an electron at an angle perpendicular to its original direction.It then scatters off a second electron such that this secondary scattered photon continues in the direction of the original photon.Calculate the difference in energy between the first and second recoiling electrons.
A)606 keV
B)194 keV
C)488 keV
D)118 keV
E)370 keV

Accepted Answer

The answer of A gamma-ray photon of energy 800 keV...

Question 3

For an electron to have a de Broglie wavelength of 0.10 nm it must be accelerated through a potential difference of
A)1.0 V
B)0.51 MV
C)0.15 kV
D)5 kV
E)0.94 kV

Accepted Answer

The answer of For an electron to have a de...

Question 4

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 answer of When a surface is illuminated with light...

Question 5

A gamma-ray photon of energy 1.00 MeV scatters off an electron at an angle perpendicular to its original direction.Calculate the kinetic energy of the recoiling electron.
A)489 keV
B)511 keV
C)663 keV
D)253 keV
E)337 keV

Accepted Answer

The Compton scattering formula is used to calculate the change in wavelength of the photon, and from there, the energy transferred to the electron. For a 90-degree scattering angle, the energy of the recoiling electron is approximately 663 keV.

Question 6

A 10.0-kg mass moving with a velocity of 100 m/s has a de Broglie wavelength of approximately A)1.98 $\times$ 10^-22 m B)1.51 $\times$ 10³⁶ m C)6.62 $\times$ 10^-33 m D)6.62 $\times$ 10^-37 m E)6.62 $\times$ 10^-31 m

Accepted Answer

The answer of A 10.0-kg mass moving with a velocity...

Question 7

In Compton scattering of X rays by electrons,the maximum shift in the X-ray wavelength occurs when the scattering angle is
A)180º
B)135º
C)90º
D)45º
E)0º

Accepted Answer

The maximum shift in the X-ray wavelength during Compton scattering occurs when the scattering angle is 180º, as this represents a complete reversal of the photon's direction, leading to the maximum possible change in momentum and hence the maximum wavelength shift.

Question 8

A gamma-ray photon is back scattered at an angle of 180° after a collision with an electron.If the scattered photon has energy 150 keV,what was the energy of the initial photon?
A)303 keV
B)363 keV
C)95 keV
D)427 keV
E)393 keV

Accepted Answer

In Compton scattering, the scattered photon's energy is given by,
E' = E/(1 + (E/Mc^2)*(1 - cos(theta))),
where E is the initial photon energy, E' is the scattered photon energy, theta is the scattering angle, M is the rest mass of the electron, and c is the speed of light.

Since the photon is backscattered (theta = 180), we have cos(theta) = -1, and solving for E we get,
E = E'/(1 - (E'/Mc^2)).

Substituting the given values, we get E = 363 keV. Therefore, the initial photon energy was 363 keV.

Hence, the correct choice is B.

Question 9

An X-ray photon of wavelength 0.10 nm is Compton-scattered from a free electron.The greatest observed shift in wavelength,as seen in the scattered photon,is
A)0.0012 nm
B)0.0024 nm
C)0.0048 nm
D)0.18 nm
E)0.10 nm

Accepted Answer

The Compton shift in wavelength is given by the formula: 
Δλ = λ - λ' = (h/mec) * (1 - cosθ)
Where λ' is the wavelength of the scattered photon, θ is the scattering angle, h is the Planck constant, me is the mass of the electron, and c is the speed of light.

The maximum shift in wavelength will occur when the scattering angle is 180° (backscattering). In this case, cosθ = -1, and the formula becomes:

Δλ = (2h/mec) = 0.0048 nm

Therefore, the greatest observed shift in wavelength is 0.0048 nm, which is choice C.

Question 10

A gamma-ray photon of energy 800 keV scatters off an electron at an angle perpendicular to its original direction.It then scatters off a second electron such that this secondary scattered photon continues in the direction of the original photon.Calculate the energy difference between the initial and final photon.
A)682 keV
B)194 keV
C)606 keV
D)118 keV
E)312 keV

Accepted Answer

The answer of A gamma-ray photon of energy 800 keV...

Question 11

The pupil area of the human eye is ~ 1 $\times$ 10^-5 m².The minimum light intensity that the eye is sensitive to is ~ 1 $\times$ 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 $\times$ 10^-2 s^-1 E)~3 $\times$ 10¹² s^-1

Accepted Answer

The answer of The pupil area of the human eye...

Question 12

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 $\times$ 10^-7 nm B)1.7 $\times$ 10^-8 m C)1.7 $\times$ 10^-9 m D)1.7 $\times$ 10^-7 m E)1.7 $\times$ 10² m

Accepted Answer

The energy of a photon is given by E = hc/λ. Rearranging for λ gives λ = hc/E. Using 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 13

The Davisson and Germer experiment verified that
A)the velocity of light in a vacuum is a universal constant.
B)electrons have an intrinsic spin.
C)waves have a particle aspect.
D)electrons show wave characteristics.
E)the neutron does not have a charge.

Accepted Answer

The Davisson and Germer experiment showed that electrons exhibit wave characteristics, specifically diffraction, when they pass through a crystal lattice. This experiment directly confirmed the wave-particle duality of electrons, which is a fundamental concept in quantum mechanics. Therefore, the correct answer is D.

Question 14

Microwaves range in wavelength from about 2.0 $\times$ 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 can be calculated using the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s), $c$ is the speed of light ($3.00 \times 10^8$ m/s), and $\lambda$ is the wavelength in meters. For the maximum energy, we use the shortest wavelength given, which is 2.0 cm or 0.02 m. Converting the energy to electron volts (1 eV = $1.602 \times 10^{-19}$ J), the calculation for the energy yields a value in the millielectronvolt (meV) range, closest to the option of 6.2 meV.

Question 15

An electron (m_e = 9.11 $\times$ 10^-31 kg)traveling with a velocity of 5.50 $\times$ 10⁶ m/s has a de Broglie wavelength of approximately A)0.0710 nm B)0.130 nm C)0.180 nm D)0.230 nm E)0.550 nm

Accepted Answer

The de Broglie wavelength for an electron is given by λ = h/p, where h is Planck's constant and p is the momentum of the electron. 
p = mv, where m is the mass of the electron and v is its velocity. 
Substituting the values given: 
p = (9.11×10^-31 kg)(5.50×10^6 m/s) = 5.01×10^-24 kg m/s 
λ = h/p = (6.626×10^-34 J s)/(5.01×10^-24 kg m/s) = 1.32×10^-10 m = 0.130 nm 
Therefore, the answer is B: 0.130 nm.

Question 16

An electron (m_e = 9.11 $\times$ 10^-31 kg)traveling with a velocity of 1.1 $\times$ 10⁶ m/s has a de Broglie wavelength of approximately A)0.66 nm B)0.78 nm C)0.89 nm D)0.97 nm E)1.2 nm

Accepted Answer

The de Broglie wavelength ($\lambda$) of a particle is given by the formula $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \text{kg} / \text{s}$), $m$ is the mass of the particle, and $v$ is the velocity of the particle. Substituting the given values ($m = 9.11 \times 10^{-31} \, \text{kg}$ and $v = 1.1 \times 10^6 \, \text{m/s}$), we can calculate the de Broglie wavelength. The calculation yields a wavelength of approximately $0.66 \, \text{nm}$, making option A the correct answer.

Question 17

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 wavelength λ, we find λ = hc/E. Substituting the given energy (10^-7 eV) and converting it to joules, we find the wavelength to be approximately 12 meters.

Question 18

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 $\times$ 10^-7 m B)2.22 $\times$ 10^-9 m C)4.50 $\times$ 10⁶ m D)4.50 $\times$ 10^-6 m E)2.22 $\times$ 10^-11 m

Accepted Answer

The threshold wavelength can be calculated using the equation $\lambda = \frac{hc}{\phi}$, where $h$ is Planck's constant (6.626 x 10$^{-34}$ J·s), $c$ is the speed of light (3.00 x 10$^8$ m/s), and $\phi$ is the work function in joules. First, convert the work function from eV to joules (1 eV = 1.602 x 10$^{-19}$ J), which gives $\phi = 5.58 \times 1.602 \times 10^{-19}$ J. Then, calculate $\lambda$, which results in a value in meters that corresponds to choice A when the correct conversion and calculations are applied.

Question 19

The work functions for two different metals A and B,are 3.23 eV and 5.33 eV,respectively.If photons of 220 nm are incident on the two metals,calculate the difference in energy of the fastest photoelectrons from metal A and metal B.
A)0.69 eV
B)0.0 eV
C)2.10 eV
D)1.72 eV
E)2.41 eV

Accepted Answer

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

E = hc/λ

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

For 220 nm, we have:

E = (6.626 x 10^-34 J.s) x (3.0 x 10^8 m/s) / (220 x 10^-9 m)

E = 9.01 x 10^-19 J

We can convert this energy to electron volts (eV) by dividing it by the fundamental charge (e):

E = 9.01 x 10^-19 J / 1.602 x 10^-19 C

E = 5.63 eV

For metal A, the maximum kinetic energy of a photoelectron is:

Kmax = E - Φ

where Φ is the work function of the metal.

Kmax = 5.63 eV - 3.23 eV = 2.40 eV

For metal B, the maximum kinetic energy of a photoelectron is:

Kmax = E - Φ

Kmax = 5.63 eV - 5.33 eV = 0.30 eV

The difference in energy of the fastest photoelectrons from metal A and metal B is:

ΔKmax = Kmax(A) - Kmax(B)

ΔKmax = 2.40 eV - 0.30 eV = 2.10 eV

Therefore, the correct answer is C) 2.10 eV.

Question 20

In a Compton scattering experiment,an 8.00-MeV incident photon is scattered at an angle of 15º with energy of 5.20 MeV.The kinetic energy of the recoil electron is
A)2.80 MeV
B)2.08 MeV
C)5.20 MeV
D)7.28 MeV
E)800 MeV

Accepted Answer

The kinetic energy of the recoil electron is the difference between the initial photon energy and the scattered photon energy: 8.00 MeV - 5.20 MeV = 2.80 MeV.

Quiz 34: Wave Particle Duality and Quantum Physics

An electron in the hydrogen atom makes a transition from an energy level of -1.51 eV to one of energy -3.40 eV.In this process a photon is emitted with what wavelength?

For an electron to have a de Broglie wavelength of 0.10 nm it must be accelerated through a potential difference of

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 gamma-ray photon of energy 1.00 MeV scatters off an electron at an angle perpendicular to its original direction.Calculate the kinetic energy of the recoiling electron.

A 10.0-kg mass moving with a velocity of 100 m/s has a de Broglie wavelength of approximately

In Compton scattering of X rays by electrons,the maximum shift in the X-ray wavelength occurs when the scattering angle is

A gamma-ray photon is back scattered at an angle of 180° after a collision with an electron.If the scattered photon has energy 150 keV,what was the energy of the initial photon?

An X-ray photon of wavelength 0.10 nm is Compton-scattered from a free electron.The greatest observed shift in wavelength,as seen in the scattered photon,is

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

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.

The Davisson and Germer experiment verified that

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

An electron (m_e = 9.11 $\times$ 10^-31 kg) traveling with a velocity of 5.50 $\times$ 10⁶ m/s has a de Broglie wavelength of approximately

An electron (m_e = 9.11 $\times$ 10^-31 kg) traveling with a velocity of 1.1 $\times$ 10⁶ m/s has a de Broglie wavelength of approximately

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

The work functions for two different metals A and B,are 3.23 eV and 5.33 eV,respectively.If photons of 220 nm are incident on the two metals,calculate the difference in energy of the fastest photoelectrons from metal A and metal B.

In a Compton scattering experiment,an 8.00-MeV incident photon is scattered at an angle of 15º with energy of 5.20 MeV.The kinetic energy of the recoil electron is