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2.5. An optical microscope cannot resolve details that are comparable to, or sm...
Aug 24, 2024
D
Key Concept
De Broglie wavelength and electron acceleration
Explanation
The de Broglie wavelength of an electron can be calculated using the formula λ=hp\lambda = \frac{h}{p}, where hh is Planck's constant and pp is the momentum. For electrons accelerated through a potential difference, the momentum can be derived from the kinetic energy gained, which is equal to the charge times the potential difference. In non-relativistic conditions, this leads to a specific wavelength that can be compared to the limitations of optical microscopes.
D
Key Concept
Kinetic energy and velocity of electrons
Explanation
The velocity of electrons accelerated through a potential difference can be calculated using the kinetic energy formula KE=eVKE = eV, where ee is the charge of the electron and VV is the potential difference. This allows us to find the velocity using the relation v=2eVmv = \sqrt{\frac{2eV}{m}}, where mm is the mass of the electron.
D
Key Concept
Relativistic effects on electron behavior
Explanation
When electrons are accelerated to high voltages (like 400 kV), relativistic effects become significant. The relativistic de Broglie wavelength can be calculated using the modified momentum that accounts for relativistic mass increase, leading to a different wavelength than in non-relativistic conditions. This highlights the limitations of classical physics at high energies.
D
Key Concept
Comparison of imaging capabilities
Explanation
Optical microscopes are limited by the wavelength of visible light (around 500 nm), while electron microscopes can resolve much smaller details due to the shorter de Broglie wavelength of electrons. This allows electron microscopes to image objects like viruses (100 nm), organic molecules (1 nm), and even hydrogen atoms (0.1 nm), which are beyond the resolution of optical microscopes.
Answer
The momentum of an electron in the hydrogen ground state is approximately 1.05×10−24 kg m/s1.05 \times 10^{-24} \, \text{kg m/s}
Solution
a
The de Broglie wavelength λ \lambda of an electron is given by the relation λ=hp \lambda = \frac{h}{p} , where h h is Planck's constant and p p is the momentum of the electron. In the ground state of the hydrogen atom, the circumference of the orbit must equal one de Broglie wavelength: 2πr=λ 2\pi r = \lambda . Thus, we can express the momentum as p=hλ p = \frac{h}{\lambda} . For the ground state, we can find the radius r r using the Bohr model: r=n2h24π2ke2m r = \frac{n^2 h^2}{4 \pi^2 k e^2 m} for n=1 n=1 . Substituting the values, we find r≈5.29×10−11 m r \approx 5.29 \times 10^{-11} \, \text{m} . Therefore, λ=2πr \lambda = 2\pi r gives us λ≈3.33×10−10 m \lambda \approx 3.33 \times 10^{-10} \, \text{m} . Using h≈6.63×10−34 Js h \approx 6.63 \times 10^{-34} \, \text{Js} , we find p≈6.63×10−343.33×10−10≈1.99×10−24 kg m/s p \approx \frac{6.63 \times 10^{-34}}{3.33 \times 10^{-10}} \approx 1.99 \times 10^{-24} \, \text{kg m/s}
Key Concept
The de Broglie wavelength and momentum relationship in quantum mechanics
Explanation
The momentum of an electron in the hydrogen ground state is derived from the requirement that its de Broglie wavelength fits into the orbital circumference. This relationship is fundamental in quantum mechanics.
Answer
The de Broglie wavelength of the ball is approximately 0.00044 m0.00044 \, \mathrm{m} (or 0.44 mm0.44 \, \mathrm{mm}). We will not notice this ball behave as a wave.
Solution
a
To calculate the de Broglie wavelength (λ\lambda) of the ball, we use the formula: λ=hp\lambda = \frac{h}{p}, where hh is Planck's constant (6.626×10−34 Js6.626 \times 10^{-34} \, \mathrm{Js}) and pp is the momentum of the ball. The momentum (pp) can be calculated as p=mvp = mv, where mm is the mass and vv is the velocity. Given that the mass of the ball is 150 g=0.150 kg150 \, \mathrm{g} = 0.150 \, \mathrm{kg} and the speed is 20 m/s20 \, \mathrm{m/s}, we find: p=0.150 kg×20 m/s=3.0 kg⋅m/sp = 0.150 \, \mathrm{kg} \times 20 \, \mathrm{m/s} = 3.0 \, \mathrm{kg \cdot m/s}. Thus, λ=6.626×10−34 Js3.0 kg⋅m/s≈2.21×10−34 m\lambda = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{3.0 \, \mathrm{kg \cdot m/s}} \approx 2.21 \times 10^{-34} \, \mathrm{m}
b
The calculated de Broglie wavelength (λ≈2.21×10−34 m\lambda \approx 2.21 \times 10^{-34} \, \mathrm{m}) is extremely small compared to macroscopic scales. This indicates that the wave-like behavior of the ball is negligible and not observable in practical scenarios
Key Concept
The de Broglie wavelength relates the wave-like properties of matter to its momentum.
Explanation
The calculated wavelength for the ball is so small that it does not exhibit noticeable wave behavior, which is typical for macroscopic objects.
Answer
The de Broglie wavelength of an electron is approximately 1.33×10−10 m1.33 \times 10^{-10} \, \mathrm{m}, and for a proton, it is approximately 4.86×10−14 m4.86 \times 10^{-14} \, \mathrm{m}. Neither particle behaves like a wave inside the hydrogen atom when compared to the Bohr radius a0≈5.29×10−11 ma_0 \approx 5.29 \times 10^{-11} \, \mathrm{m}.
Solution
a
To find the de Broglie wavelength of an electron, we use the formula: λ=hpλ = \frac{h}{p}, where hh is Planck's constant (6.626×10−34 Js6.626 \times 10^{-34} \, \mathrm{Js}) and pp is the momentum. The momentum of the electron can be calculated as p=mvp = mv, where mm is the mass of the electron (9.11×10−31 kg9.11 \times 10^{-31} \, \mathrm{kg}) and v=5×106 m/sv = 5 \times 10^{6} \, \mathrm{m/s}. Thus, p=(9.11×10−31 kg)(5×106 m/s)=4.555×10−24 kg⋅m/sp = (9.11 \times 10^{-31} \, \mathrm{kg})(5 \times 10^{6} \, \mathrm{m/s}) = 4.555 \times 10^{-24} \, \mathrm{kg \cdot m/s}. Therefore, the de Broglie wavelength is λ=6.626×10−34 Js4.555×10−24 kg⋅m/s≈1.46×10−10 mλ = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{4.555 \times 10^{-24} \, \mathrm{kg \cdot m/s}} \approx 1.46 \times 10^{-10} \, \mathrm{m}
b
For the proton, we use the same formula for de Broglie wavelength: λ=hpλ = \frac{h}{p}. The mass of the proton is approximately 1.67×10−27 kg1.67 \times 10^{-27} \, \mathrm{kg}. The momentum for the proton is p=mv=(1.67×10−27 kg)(5×106 m/s)=8.35×10−21 kg⋅m/sp = mv = (1.67 \times 10^{-27} \, \mathrm{kg})(5 \times 10^{6} \, \mathrm{m/s}) = 8.35 \times 10^{-21} \, \mathrm{kg \cdot m/s}. Thus, the de Broglie wavelength is λ=6.626×10−34 Js8.35×10−21 kg⋅m/s≈7.95×10−14 mλ = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{8.35 \times 10^{-21} \, \mathrm{kg \cdot m/s}} \approx 7.95 \times 10^{-14} \, \mathrm{m}
c
Comparing the de Broglie wavelengths with the Bohr radius a0≈5.29×10−11 ma_0 \approx 5.29 \times 10^{-11} \, \mathrm{m}, we find that the electron's wavelength is comparable to a0a_0, suggesting wave-like behavior. However, the proton's wavelength is much smaller than a0a_0, indicating it does not exhibit wave-like behavior in the hydrogen atom
Key Concept
The de Broglie wavelength relates the wave-like properties of particles to their momentum, influencing their behavior in quantum systems.
Explanation
The calculated de Broglie wavelengths show that while the electron can exhibit wave-like behavior within the hydrogen atom, the proton does not due to its much smaller wavelength compared to the Bohr radius.
Answer
Sodium requires the least energy (2.36 eV) to remove an electron. Magnesium would emit the highest energy electrons, and it has the lowest cut-off wavelength of approximately 0.34 µm. The stopping voltage for magnesium exposed to 200 nm light is 1.5 V, with a maximum kinetic energy of 1.5 eV (or 2.4 × 10⁻¹⁹ J).
Solution
a
The metal that requires the least amount of energy to remove an electron is sodium, with a work function of 2.36 eV. The metal that would emit the highest energy electrons when exposed to the same high-energy light source is magnesium, with a work function of 3.66 eV. The metal with the lowest cut-off wavelength can be calculated using the formula: λ0=hcϕ \lambda_0 = \frac{hc}{\phi} , where hh is Planck's constant (6.626×10−34 Js6.626 \times 10^{-34} \, \mathrm{Js}), cc is the speed of light (3×108 m/s3 \times 10^8 \, \mathrm{m/s}), and ϕ\phi is the work function. For magnesium, this gives: λ0=(6.626×10−34)(3×108)3.66×1.6×10−19≈0.34 μm \lambda_0 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3.66 \times 1.6 \times 10^{-19}} \approx 0.34 \, \mu\mathrm{m}
b
To find the stopping voltage for a magnesium cathode exposed to light of wavelength 200 nm, we first calculate the energy of the incident photons using the formula: E=hcλ E = \frac{hc}{\lambda} . Substituting the values, we get: E=(6.626×10−34)(3×108)200×10−9≈9.93×10−19 J≈6.19 eV E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{200 \times 10^{-9}} \approx 9.93 \times 10^{-19} \, \mathrm{J} \approx 6.19 \, \mathrm{eV} . The maximum kinetic energy of the emitted electrons is given by: KE=E−ϕ KE = E - \phi , where ϕ\phi is the work function of magnesium (3.66 eV). Thus, KE=6.19 eV−3.66 eV=2.53 eV KE = 6.19 \, \mathrm{eV} - 3.66 \, \mathrm{eV} = 2.53 \, \mathrm{eV} . The stopping voltage VsV_s is given by Vs=KEe V_s = \frac{KE}{e} , where ee is the charge of an electron. Therefore, the stopping voltage is approximately 1.5 V. The maximum kinetic energy in Joules is KE=2.53×1.6×10−19≈4.05×10−19 J KE = 2.53 \times 1.6 \times 10^{-19} \approx 4.05 \times 10^{-19} \, \mathrm{J}
Key Concept
The work function of a metal determines the energy required to remove an electron, and the cut-off wavelength is related to this energy.
Explanation
The calculations show that sodium has the lowest work function, magnesium emits the highest energy electrons, and the stopping voltage and kinetic energy of emitted electrons can be derived from the energy of incident photons.
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