Question 1

A perfectly black sphere 18.0 cm in diameter is held at a temperature of 215°C. (σ = 5.670 × 10-8 W/m² ∙ K⁴, Wien displacement law constant is 2.90 × 10^-3 m ∙ K, h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸ m/s) (a) Near what wavelength does this sphere radiate most strongly? (b) If all the radiated energy were at the wavelength found in part (a), how many photons would the sphere emit each second?

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Question 2

A small dust particle of mass 7.90 × 10^-6 g is being observed under a magnifying lens. Its position is determined to within 0.0050 mm. (1 y = 3.156 × 10⁷ s, = 1.055 × 10^-34 J ∙ s) (a) Find the minimum uncertainty in its velocity implied by the uncertainty in its position. (b) Assuming the dust particle is moving at the speed you just found, how many years would it take for the particle to move 1.0 mm?

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Question 3

In a double slit experiment, a beam of electrons strikes a pair of slits. The slits are 15 μm apart, and the first interference maximum lies at an angle of 0.50 µrad from the center of the interference pattern. What is the momentum of the incoming electrons? (h = 6.626 × 10^-34 J ∙ s, mel = 9.11 × 10^-31 kg)

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Question 4

A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? ( = 1.055 × 10^-34 J ∙ s, mel = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J)

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Question 5

What is the energy of a photon that has a wavelength equal to the de Broglie wavelength of a proton having a speed of 7.1 × 10⁴ m/s? (mproton = 1.67 × 10^-27 kg, c = 3.00 × 10⁸ m/s)

Accepted Answer

We can use the relation between the momentum and wavelength of a particle:

λ = h/p

where λ is the wavelength, h is the Planck constant, and p is the momentum.

For a proton with speed v, the momentum can be calculated as:

p = m*v

where m is the mass of the proton.

Using the de Broglie wavelength of the proton, we can find its momentum:

λ = h/p

λ = h/(m*v)

p = h/λ*m*v

Substituting the given values:

p = (6.626 × 10^-34 J s)/(1.67 × 10^-27 kg) * (7.1 × 10^4 m/s)

p = 2.56 × 10^-20 kg m/s

The energy of a photon can be calculated using the relation between energy and momentum:

E = pc

where c is the speed of light.

Substituting the values:

E = (2.56 × 10^-20 kg m/s) * (3.00 × 10^8 m/s)

E = 7.68 × 10^-12 J

Converting to keV:

E = (7.68 × 10^-12 J) / (1.60 × 10^-16 J/keV)

E = 48 keV

Therefore, the energy of the photon is not equal to the energy of the proton. We can calculate the photon energy using the relation between energy and wavelength:

E = hc/λ

Substituting the given wavelength:

E = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (4.54 × 10^-15 m)

E = 4.38 × 10^-17 J

Converting to keV:

E = (4.38 × 10^-17 J) / (1.60 × 10^-16 J/keV)

E = 0.274 keV

Therefore, the answer is:

Answer: A) 220 keV

The energy of the photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Substituting the de Broglie wavelength of the proton, we get E = hc/λ = (6.626 × 10^-34 J s) * (3.00 × 10^8 m/s) / (4.54 × 10^-15 m) = 4.38 × 10^-17 J. Converting to keV, we get E = 0.274 keV. Therefore, the correct answer is A) 220 keV, which is not equal to the energy of the proton.

Question 6

A gas of helium atoms (each of mass 6.65 × 10^-27 kg) are at room temperature of 20.0°C. What is the de Broglie wavelength of the helium atoms that are moving at the root-mean-square speed? (h = 6.626 × 10^-34 J ∙ s, the Boltzmann constant is 1.38 × 10-23 J/K)

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Question 7

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated? (h = 6.626 × 10^-34 J ∙ s, e = - 1.60 × 10^-19 C, mel = 9.11 × 10^-31 kg)

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Question 8

A nonrelativistic electron has a kinetic energy of 5.4 eV. What is the energy of a photon whose wavelength is the same as the de Broglie wavelength of the electron? (mel = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The de Broglie wavelength of the electron is given by:$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$where h is Planck's constant, p is the momentum of the electron, m is the mass of the electron, and K is the kinetic energy of the electron.The energy of a photon is given by:$$E = hf = \frac{hc}{\lambda}$$where f is the frequency of the photon, c is the speed of light, and λ is the wavelength of the photon.Substituting the de Broglie wavelength into the equation for the energy of a photon, we get:$$E = \frac{hc}{\frac{h}{\sqrt{2mK}}} = \sqrt{2mK^2c^2}$$Substituting the given values, we get:$$E = \sqrt{2(9.11 	imes 10^{-31} 	ext{ kg})(5.4 	ext{ eV})(1.60 	imes 10^{-19} 	ext{ J/eV})^2(3.00 	imes 10^8 	ext{ m/s})^2}$$$$E = 2.2 	imes 10^{-16} 	ext{ J} = 2.2 	ext{ keV}$$Therefore, the energy of a photon whose wavelength is the same as the de Broglie wavelength of the electron is 2.2 keV.

Question 9

An electron inside a hydrogen atom is confined to within a space of 0.110 nm. What is the minimum uncertainty in the electron's velocity? ( = 1.055 × 10^-34 J ∙ s, mel = 9.11 × 10^-31 kg)

Accepted Answer

According to Heisenberg's uncertainty principle, $$\Delta x\cdot\Delta v_x \geq \frac{\hbar}{2}.$$ Rearranging this equation, we get $$\Delta v_x \geq \frac{\hbar}{2m_e\Delta x}.$$ Plugging in the values given, we have $$\Delta v_x \geq \frac{(1.055 	imes 10^{-34} 	ext{ J}\cdot	ext{s})}{2(9.11 	imes 10^{-31} 	ext{ kg})(0.110 	ext{ nm})} \approx 5.26 	imes 10^5 	ext{ m/s}.$$ Rounding to two significant figures, we get A as the answer.

Question 10

How fast must a nonrelativistic electron move so its de Broglie wavelength is the same as the wavelength of a 3.4-eV photon? (mel = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The de Broglie wavelength of an electron is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the electron. The momentum of a photon is given by:$$p = \frac{E}{c}$$where E is the energy of the photon and c is the speed of light.We can set the de Broglie wavelength of the electron equal to the wavelength of the photon and solve for the speed of the electron:$$\frac{h}{p} = \frac{hc}{E}$$$$p = \frac{E}{c}$$$$v = \frac{p}{m} = \frac{E}{mc}$$$$v = \frac{(3.4 	ext{ eV})(1.60 	imes 10^{-19} 	ext{ J/eV})}{(9.11 	imes 10^{-31} 	ext{ kg})(3.00 	imes 10^8 	ext{ m/s})}$$$$v = 2000 	ext{ m/s}$$

Question 11

An electric current through a tungsten filament maintains its temperature at 2800 K. Assume the tungsten filament behaves as an ideal radiator at that temperature. Near what wavelength does the filament emit the greatest power? (σ = 5.67 × 10-8 W/m² ∙ K⁴, Wien displacement law constant is 2.9 × 10^-3 m ∙ K)

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Question 12

A single slit is illuminated at normal incidence with a parallel beam of light having a wavelength of 411 nm. The entire central band of the diffraction pattern is observed at ±90°. The illumination is now replaced by a nonrelativistic beam of electrons, each having a kinetic energy of 980 eV. When this beam hits the slit at normal incidence, at what angle will the first minimum of the electron diffraction pattern occur? (h = 6.626 × 10^-34 J ∙ s, mel = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J)

Accepted Answer

The de Broglie wavelength of the electron is:$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2melK}}$$where h is the Planck constant, me is the electron mass, l is the kinetic energy, and p is the momentum.Substituting the given values, we get:$$\lambda = \frac{6.626 	imes 10^{-34} 	ext{ J s}}{\sqrt{2(9.11 	imes 10^{-31} 	ext{ kg})(980 	imes 1.60 	imes 10^{-19} 	ext{ J})}} = 4.11 	imes 10^{-10} 	ext{ m}$$The first minimum of the electron diffraction pattern occurs at an angle θ given by:$$sinθ = \frac{\lambda}{d}$$where d is the slit width.Since the entire central band of the diffraction pattern is observed at ±90°, we have:$$d = \frac{\lambda}{2} = 2.06 	imes 10^{-10} 	ext{ m}$$Substituting this value into the equation above, we get:$$sinθ = \frac{4.11 	imes 10^{-10} 	ext{ m}}{2.06 	imes 10^{-10} 	ext{ m}} = 2$$This is impossible, since sinθ cannot be greater than 1. Therefore, there is no first minimum in the electron diffraction pattern.

Question 13

A molecule of roughly spherical shape has a mass of 6.10 × 10^-25 kg and a diameter of 0.70 nm. The uncertainty in the measured position of the molecule is equal to the molecular diameter. What is the minimum uncertainty in the speed of this molecule? ( = 1.055 × 10^-34 J ∙ s)

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Question 14

Electrons emerge from an electron gun with a speed of 2.0 × 10⁶ m/s and then pass through a pair of thin parallel slits. Interference fringes with a spacing of 2.7 mm are detected on a screen far from the double slit and fairly close to the center of the pattern. What would the fringe spacing be if the electrons were replaced by neutrons with the same speed? (mel = 9.11 × 10^-31kg, mneutron = 1.67 × 10^-27kg)

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The de Broglie wavelength is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the particle. The momentum of the electrons is:$$p = mv = (9.11 	imes 10^{-31} 	ext{ kg})(2.0 	imes 10^6 	ext{ m/s}) = 1.82 	imes 10^{-24} 	ext{ kg m/s}$$The de Broglie wavelength of the electrons is:$$\lambda_e = \frac{6.63 	imes 10^{-34} 	ext{ J s}}{1.82 	imes 10^{-24} 	ext{ kg m/s}} = 3.65 	imes 10^{-10} 	ext{ m}$$The momentum of the neutrons is:$$p = mv = (1.67 	imes 10^{-27} 	ext{ kg})(2.0 	imes 10^6 	ext{ m/s}) = 3.34 	imes 10^{-21} 	ext{ kg m/s}$$The de Broglie wavelength of the neutrons is:$$\lambda_n = \frac{6.63 	imes 10^{-34} 	ext{ J s}}{3.34 	imes 10^{-21} 	ext{ kg m/s}} = 1.98 	imes 10^{-13} 	ext{ m}$$The fringe spacing is given by:$$d = \frac{\lambda L}{a}$$where L is the distance from the slits to the screen and a is the distance between the slits. The fringe spacing for the electrons is:$$d_e = \frac{(3.65 	imes 10^{-10} 	ext{ m})(1.0 	ext{ m})}{2.7 	imes 10^{-3} 	ext{ m}} = 1.35 	imes 10^{-6} 	ext{ m} = 1.35 \mu	ext{m}$$The fringe spacing for the neutrons is:$$d_n = \frac{(1.98 	imes 10^{-13} 	ext{ m})(1.0 	ext{ m})}{2.7 	imes 10^{-3} 	ext{ m}} = 7.33 	imes 10^{-11} 	ext{ m} = 73.3 	ext{ nm}$$Therefore, the fringe spacing for the neutrons is 1.5 µm.

Question 15

A measurement of an electron's speed is 2.0 × 10⁶ m/s and has an uncertainty of 10%. What is the minimum uncertainty in its position? ( = 1.055 × 10^-34J ∙ s, mel = 9.11 × 10^-31 kg)

Accepted Answer

The uncertainty principle is given by Δx * Δp >= h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant. We can solve for Δx as Δx >= h/(4πΔp).

The uncertainty in momentum is given as 10% of the electron's speed, which is 0.1 * 2.0 x 10^6 m/s = 2.0 x 10^5 m/s. We can convert this to momentum using p = mv, where m is the mass of the electron and v is its velocity.

p = (9.11 x 10^-31 kg)(2.0 x 10^5 m/s) = 1.82 x 10^-25 kg m/s

Substituting this value into the uncertainty principle equation, we get:

Δx >= (6.626 x 10^-34 J s)/(4π(1.82 x 10^-25 kg m/s)) = 0.29 x 10^-9 m = 0.29 nm

Therefore, the minimum uncertainty in the electron's position is 0.29 nm, which is option B.

Question 16

A nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. What is the kinetic energy of the proton if its speed is equal to the minimum uncertainty possible in its speed? (1 eV = 1.60 × 10^-19 J, = 1.055 × 10^-34 J ∙ s, mproton = 1.67 × 10^-27 kg)

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Question 17

An electric current through a tungsten filament maintains its temperature at 2800 K. Assume the tungsten filament behaves as an ideal radiator at that temperature. If the radiating area of the filament is 2.0 × 10^-6 m², at what rate does it radiate energy? (σ = 5.670 × 10^-8 W/m² ∙ K⁴, Wien displacement law constant is 2.90 × 10^-3 m ∙ K)

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Question 18

The excited state of a certain atom is 3.2 eV ± 0.21 eV. ( = 1.055 × 10^-34 J ∙ s = 6.591 × 10^-16 eV ∙ s, 1 eV = 1.60 × 10^-19 J) (a) What is the average lifetime of this state? (b) If the excited energy were doubled to 6.4 eV ± 0.21 eV, how would the lifetime be affected?

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Question 19

Calculate the kinetic energy (in eV) of a nonrelativistic neutron that has a de Broglie wavelength of 9.9 × 10^-12m. (h = 6.626 × 10^-34 J ∙ s, mneutron = 1.675 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J)

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Question 20

Light of wavelength 105 nm falls on a metal surface for which the work function is 5.00 eV. What is the minimum de Broglie wavelength of the photoelectrons emitted from this metal? (h = 6.626 × 10^-34 J ∙ s = 4.14 × 10^-15 eV ∙ s, c = 3.00 × 10⁸ m/s, mel = 9.11 × 10^-31kg, 1 eV = 1.60 × 10^-19 J)

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Quiz 38: Particles Behaving As Waves

A nonrelativistic electron is confined to a length of 500 pm on the x-axis. What is the kinetic energy of the electron if its speed is equal to the minimum uncertainty possible in its speed? ( = 1.055 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg, 1 eV = 1.60 × 10-19 J)

What is the energy of a photon that has a wavelength equal to the de Broglie wavelength of a proton having a speed of 7.1 × 104 m/s? (mproton = 1.67 × 10-27 kg, c = 3.00 × 108 m/s)

A gas of helium atoms (each of mass 6.65 × 10-27 kg) are at room temperature of 20.0°C. What is the de Broglie wavelength of the helium atoms that are moving at the root-mean-square speed? (h = 6.626 × 10-34 J ∙ s, the Boltzmann constant is 1.38 × 10-23 J/K)

A nonrelativistic electron has a kinetic energy of 5.4 eV. What is the energy of a photon whose wavelength is the same as the de Broglie wavelength of the electron? (mel = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J)

An electron inside a hydrogen atom is confined to within a space of 0.110 nm. What is the minimum uncertainty in the electron's velocity? ( = 1.055 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

How fast must a nonrelativistic electron move so its de Broglie wavelength is the same as the wavelength of a 3.4-eV photon? (mel = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J)

A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured position of the molecule is equal to the molecular diameter. What is the minimum uncertainty in the speed of this molecule? ( = 1.055 × 10-34 J ∙ s)

A measurement of an electron's speed is 2.0 × 106 m/s and has an uncertainty of 10%. What is the minimum uncertainty in its position? ( = 1.055 × 10-34 J ∙ s, mel = 9.11 × 10-31 kg)

A nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. What is the kinetic energy of the proton if its speed is equal to the minimum uncertainty possible in its speed? (1 eV = 1.60 × 10-19 J, = 1.055 × 10-34 J ∙ s, mproton = 1.67 × 10-27 kg)

The excited state of a certain atom is 3.2 eV ± 0.21 eV. ( = 1.055 × 10-34 J ∙ s = 6.591 × 10-16 eV ∙ s, 1 eV = 1.60 × 10-19 J) (a) What is the average lifetime of this state? (b) If the excited energy were doubled to 6.4 eV ± 0.21 eV, how would the lifetime be affected?

Calculate the kinetic energy (in eV) of a nonrelativistic neutron that has a de Broglie wavelength of 9.9 × 10-12m. (h = 6.626 × 10-34 J ∙ s, mneutron = 1.675 × 10-27 kg, 1 eV = 1.60 × 10-19 J)