Question 1

The Hubble Space Telescope has an orbital speed of 7.56 × 10³ m/s and a mass of 11 600 kg.What is the de Broglie wavelength of the telescope?

Accepted Answer

The answer of The Hubble Space Telescope has an orbital...

Question 2

What happens to the de Broglie wavelength of an electron if its momentum is reduced to one-half of its initial value?

Accepted Answer

The de Broglie wavelength is inversely proportional to the momentum of the electron. Therefore, if the momentum is reduced to one-half of its initial value, the wavelength will increase by a factor of 2 (because the momentum is reduced by a factor of 2).

Question 3

In an experiment to determine the speed and position of an electron (m = 9.11 × 10^-³¹ kg),three researchers claim to have measured the position of the electron to within ± 10^-⁹ m.They reported the following values for the speed of the electron: Which of these measurements violates one or more basic laws of modern physics?

Accepted Answer

Option A violates the Heisenberg uncertainty principle which states that it is impossible to determine both the position and momentum of a particle with absolute precision. The product of the uncertainties in position and momentum must be greater than or equal to Planck's constant.

Option B violates the speed limit given by Einstein's theory of relativity, which states that no object can travel faster than the speed of light in a vacuum.

Option C violates both the uncertainty principle and the speed limit given by Einstein's theory of relativity. Therefore, option C is the correct answer.

Question 4

The speed of a bullet with a mass of 0.050 kg is 420 m/s with an uncertainty of 0.010 %.What is the minimum uncertainty in the bullet's position if it is measured at the same time as the speed is measured?

Accepted Answer

The minimum uncertainty in the bullet's position can be calculated using the Heisenberg Uncertainty Principle, which is given by ΔxΔp ≥ ħ/2, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and ħ is the reduced Planck's constant (approximately 1.0545718 × 10^-34 m^2 kg / s). The uncertainty in momentum (Δp) can be calculated from the uncertainty in velocity (Δv) since Δp = mΔv, where m is the mass of the bullet. Given that the uncertainty in velocity is 0.010% of 420 m/s, Δv = 0.001 * 420 m/s = 0.42 m/s. Therefore, Δp = 0.050 kg * 0.42 m/s = 0.021 kg m/s. Using the Heisenberg Uncertainty Principle, Δx ≥ ħ / (2Δp) = (1.0545718 × 10^-34 m^2 kg / s) / (2 * 0.021 kg m/s) ≈ 2.5 × 10^-32 m, which corresponds to choice B.

Question 5

The position of a hydrogen atom (m = 1.7 × 10^-²⁷ kg)is known to within 2.0 × 10^-⁶ m.What is the minimum uncertainty in the atom's velocity?

Accepted Answer

The uncertainty principle states Δx * Δv ≥ ħ/(2m), where Δx is position uncertainty, Δv is velocity uncertainty, ħ is reduced Planck's constant (1.0545718 × 10^-34 Js), and m is mass. Solving for Δv gives approximately 0.016 m/s.

Question 6

The position of a 1-g object moving in the x direction at 1 cm/s is known to within ± 10 nm.In which range is the fractional uncertainty, ,in the x component of its momentum?

Accepted Answer

The answer of The position of a 1-g object moving...

Question 7

A beam of electrons is incident on a single slit that has a width of 1.00 × 10^-⁶ m and a diffraction pattern is observed on a screen located 15.0 m from the slit.The momentum of an individual electron in the beam is 5.91 × 10^-²⁴ kg × m/s.What is the width of the central maximum fringe?

Accepted Answer

The answer of A beam of electrons is incident on...

Question 8

If Planck's constant were changed to 660 J × s,what would be the minimum uncertainty in the position of a 120-kg football player running at a speed of 3.5 m/s?

Accepted Answer

The answer of If Planck's constant were changed to 660...

Question 9

The x component of the velocity of an electron (m = 9.11 × 10^-³¹ kg)is known to be between 100 m/s and 300 m/s.Which one of the following is a true statement concerning the uncertainty in the x coordinate of the electron?

Accepted Answer

The uncertainty principle, formulated by Werner Heisenberg, can be used to calculate the minimum uncertainty in position ($\Delta x$) given the uncertainty in momentum ($\Delta p$). The principle is given by $\Delta x \Delta p \geq \frac{\hbar}{2}$, where $\hbar$ is the reduced Planck's constant ($\approx 1.0545718 \times 10^{-34} m^2 kg / s$). The uncertainty in the momentum can be calculated using the range of velocities given (100 m/s to 300 m/s) and the mass of the electron ($9.11 \times 10^{-31} kg$). The uncertainty in velocity ($\Delta v$) is $300 m/s - 100 m/s = 200 m/s$, so $\Delta p = m \Delta v = 9.11 \times 10^{-31} kg \times 200 m/s = 1.822 \times 10^{-28} kg \cdot m/s$. Using the uncertainty principle:$$\Delta x \geq \frac{\hbar}{2\Delta p} \approx \frac{1.0545718 \times 10^{-34} m^2 kg / s}{2 \times 1.822 \times 10^{-28} kg \cdot m/s} \approx \frac{1.0545718 \times 10^{-34}}{3.644 \times 10^{-28}} m \approx 2.89 \times 10^{-7} m$$This calculation shows that the minimum uncertainty in the x coordinate of the electron is about $3 \times 10^{-7} m$ or $3 \times 10^{-7}$ meters, which corresponds to choice B.

Question 10

It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm.The mass of an electron is 9.11 × 10S1U1P1-S1S1P0³¹ kg.
-Suppose it is desired to observe diffraction effects for a beam of electromagnetic radiation using the same grating.Roughly,what is the required energy of the individual photons in the beam?

Accepted Answer

The energy of photons required to observe diffraction effects is related to their wavelength, which should be comparable to the grating spacing (10 nm). Using the energy-wavelength relation for photons (E = hc/λ), the energy is approximately 10² eV.

Question 11

In a computer monitor,electrons approach the screen at 1.20 × 10⁸ m/s.What is the de Broglie wavelength of these electrons? Note: the mass of electrons is 9.109 × 10^-³¹ kg;and use the relativistic momentum in your calculation.

Accepted Answer

The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant ($6.626 \times 10^{-34}$ m$^2$kg/s) and $p$ is the relativistic momentum. The relativistic momentum can be calculated using the formula $p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $m$ is the mass of the electron ($9.109 \times 10^{-31}$ kg), $v$ is the velocity of the electron ($1.20 \times 10^8$ m/s), and $c$ is the speed of light ($3.00 \times 10^8$ m/s). After calculating $p$, substitute it into the de Broglie wavelength formula to find $\lambda$. The correct calculation leads to the wavelength closest to the value given in option B.

Question 12

A proton (mp = 1.673 × 10^-²⁷ kg)and an electron (mₑ = 9.109 × 10^-³¹ kg)are confined such that the x position of each is known to within 1.50 × 10^-¹⁰ m.What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton,?

Accepted Answer

The answer of A proton (mp = 1.673 × 10^-²⁷...

Question 13

What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10^-³¹ kg;and the neutron mass is 1.67 × 10^-²⁷ kg.

Accepted Answer

The de Broglie wavelength of a particle is given by:$$\lambda = \frac{h}{p}$$where h is Planck's constant and p is the momentum of the particle. The kinetic energy of a particle is given by:$$K = \frac{p^2}{2m}$$where m is the mass of the particle. We can combine these two equations to get:$$\lambda = \frac{h}{\sqrt{2mK}}$$If the diffraction pattern of a beam of electrons is similar to that of a beam of 1.00 eV neutrons, then the de Broglie wavelength of the electrons must be the same as the de Broglie wavelength of the neutrons. Therefore, we can set the two equations equal to each other and solve for K:$$\frac{h}{\sqrt{2m_eK}} = \frac{h}{\sqrt{2m_n(1.00 	ext{ eV})}}$$where me is the mass of the electron and mn is the mass of the neutron. Solving for K, we get:$$K = \frac{m_e}{m_n}(1.00 	ext{ eV}) = (9.11 	imes 10^{-31} 	ext{ kg})/(1.67 	imes 10^{-27} 	ext{ kg})(1.00 	ext{ eV}) = 1830 	ext{ eV}$$Therefore, the kinetic energy of each electron in the beam is 1830 eV.

Question 14

It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm.The mass of an electron is 9.11 × 10S1U1P1-S1S1P0³¹ kg.
-What is the approximate kinetic energy of electrons that would be diffracted by such a grating?

Accepted Answer

To observe diffraction, the de Broglie wavelength of the electrons should be comparable to the spacing of the grating, which is 10 nm. The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{s}^{-1}$) and $p$ is the momentum of the electron. The momentum $p$ can be related to kinetic energy $E$ through the relation $p = \sqrt{2mE}$, where $m$ is the mass of the electron. Substituting the given values and solving for $E$ gives a kinetic energy in the order of $10^2$ eV, making option C the closest match.

Quiz 29: Particles and Waves

The Hubble Space Telescope has an orbital speed of 7.56 × 10³ m/s and a mass of 11 600 kg.What is the de Broglie wavelength of the telescope?

What happens to the de Broglie wavelength of an electron if its momentum is reduced to one-half of its initial value?

The speed of a bullet with a mass of 0.050 kg is 420 m/s with an uncertainty of 0.010 %.What is the minimum uncertainty in the bullet's position if it is measured at the same time as the speed is measured?

The position of a hydrogen atom (m = 1.7 × 10^-²⁷ kg) is known to within 2.0 × 10^-⁶ m.What is the minimum uncertainty in the atom's velocity?

A beam of electrons is incident on a single slit that has a width of 1.00 × 10^-⁶ m and a diffraction pattern is observed on a screen located 15.0 m from the slit.The momentum of an individual electron in the beam is 5.91 × 10^-²⁴ kg × m/s.What is the width of the central maximum fringe?

If Planck's constant were changed to 660 J × s,what would be the minimum uncertainty in the position of a 120-kg football player running at a speed of 3.5 m/s?

The x component of the velocity of an electron (m = 9.11 × 10^-³¹ kg) is known to be between 100 m/s and 300 m/s.Which one of the following is a true statement concerning the uncertainty in the x coordinate of the electron?

In a computer monitor,electrons approach the screen at 1.20 × 10⁸ m/s.What is the de Broglie wavelength of these electrons? Note: the mass of electrons is 9.109 × 10^-³¹ kg;and use the relativistic momentum in your calculation.

A proton (mp = 1.673 × 10^-²⁷ kg) and an electron (mₑ = 9.109 × 10^-³¹ kg) are confined such that the x position of each is known to within 1.50 × 10^-¹⁰ m.What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton,?

What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10^-³¹ kg;and the neutron mass is 1.67 × 10^-²⁷ kg.

It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm.The mass of an electron is 9.11 × 10^-³¹ kg. -What is the approximate kinetic energy of electrons that would be diffracted by such a grating?

Quiz 29: Particles and Waves

The Hubble Space Telescope has an orbital speed of 7.56 × 10³ m/s and a mass of 11 600 kg.What is the de Broglie wavelength of the telescope?

What happens to the de Broglie wavelength of an electron if its momentum is reduced to one-half of its initial value?

The speed of a bullet with a mass of 0.050 kg is 420 m/s with an uncertainty of 0.010 %.What is the minimum uncertainty in the bullet's position if it is measured at the same time as the speed is measured?

The position of a hydrogen atom (m = 1.7 × 10-²⁷ kg) is known to within 2.0 × 10-⁶ m.What is the minimum uncertainty in the atom's velocity?

The position of a 1-g object moving in the x direction at 1 cm/s is known to within ± 10 nm.In which range is the fractional uncertainty, ,in the x component of its momentum?

A beam of electrons is incident on a single slit that has a width of 1.00 × 10-⁶ m and a diffraction pattern is observed on a screen located 15.0 m from the slit.The momentum of an individual electron in the beam is 5.91 × 10-²⁴ kg × m/s.What is the width of the central maximum fringe?

If Planck's constant were changed to 660 J × s,what would be the minimum uncertainty in the position of a 120-kg football player running at a speed of 3.5 m/s?

The x component of the velocity of an electron (m = 9.11 × 10-³¹ kg) is known to be between 100 m/s and 300 m/s.Which one of the following is a true statement concerning the uncertainty in the x coordinate of the electron?

In a computer monitor,electrons approach the screen at 1.20 × 10⁸ m/s.What is the de Broglie wavelength of these electrons? Note: the mass of electrons is 9.109 × 10-³¹ kg;and use the relativistic momentum in your calculation.

A proton (mp = 1.673 × 10-²⁷ kg) and an electron (mₑ = 9.109 × 10-³¹ kg) are confined such that the x position of each is known to within 1.50 × 10-¹⁰ m.What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton,?

What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10-³¹ kg;and the neutron mass is 1.67 × 10-²⁷ kg.

It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm.The mass of an electron is 9.11 × 10-³¹ kg. -What is the approximate kinetic energy of electrons that would be diffracted by such a grating?

The position of a hydrogen atom (m = 1.7 × 10^-²⁷ kg) is known to within 2.0 × 10^-⁶ m.What is the minimum uncertainty in the atom's velocity?

A beam of electrons is incident on a single slit that has a width of 1.00 × 10^-⁶ m and a diffraction pattern is observed on a screen located 15.0 m from the slit.The momentum of an individual electron in the beam is 5.91 × 10^-²⁴ kg × m/s.What is the width of the central maximum fringe?

The x component of the velocity of an electron (m = 9.11 × 10^-³¹ kg) is known to be between 100 m/s and 300 m/s.Which one of the following is a true statement concerning the uncertainty in the x coordinate of the electron?

In a computer monitor,electrons approach the screen at 1.20 × 10⁸ m/s.What is the de Broglie wavelength of these electrons? Note: the mass of electrons is 9.109 × 10^-³¹ kg;and use the relativistic momentum in your calculation.

A proton (mp = 1.673 × 10^-²⁷ kg) and an electron (mₑ = 9.109 × 10^-³¹ kg) are confined such that the x position of each is known to within 1.50 × 10^-¹⁰ m.What is the ratio of the minimum uncertainty in the x component of the velocity of the electron to that of the proton,?

What is the kinetic energy of each electron in a beam of electrons if the beam produces a diffraction pattern of a crystal which is similar to that of a beam of 1.00 eV neutrons? Note: The electron mass is 9.11 × 10^-³¹ kg;and the neutron mass is 1.67 × 10^-²⁷ kg.

It is desired to obtain a diffraction pattern for electrons using a diffraction grating with lines separated by 10 nm.The mass of an electron is 9.11 × 10^-³¹ kg. -What is the approximate kinetic energy of electrons that would be diffracted by such a grating?