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Max Speed of Electrons

बनाया गया: 6 जनवरी 2025

का उपयोग करके उत्तर दिया गया O1 द्वारा Chat01

प्रश्न

Yellow light of wavelength 585 nm falls on a photocathode whose threshold frequency is
4.10 x 1014 Hz. What is the maximum speed of ejected electrons? What happens to the speed
of electrons when the intensity of the yellow light increases? Why?

विचार

Calculating frequency

I started by converting the given wavelength of 585 nm to meters and then determined the frequency using the speed of light formula, resulting in approximately 5.13 x 10^14 Hz.

Comparing energies

I calculated the photon energies at both the incident and threshold frequencies. The difference in energy is about 6.9 x 10^-20 J, indicating the kinetic energy of ejected electrons.

Calculating velocity

I determined the velocity of the ejected electrons using the square root of twice the kinetic energy divided by the electron mass, resulting in approximately 1.23 x 10^5 m/s.

Refining calculations

I recalculated the threshold energy using Planck's constant and threshold frequency, confirming it to be approximately 1.70 eV. This aligns with previous values obtained.

Calculating electron velocity

I determined the kinetic energy of electrons by subtracting the threshold energy of 1.70 eV from the photon energy of approximately 2.12 eV, resulting in a velocity of about 1.2 x 10^5 m/s.

Thought about photoelectric effect and intensity，持续 17 秒

उत्तर

Solution Outline

  1. Identify the known quantities

    • Wavelength of incident (yellow) light, λ=585  nm=585×10−9 m\lambda = 585\;\text{nm} = 585\times 10^{-9}\,\text{m}.
    • Speed of light, c=3.00×108 m/sc = 3.00 \times 10^8\,\text{m/s}.
    • Planck’s constant, h=6.626×10−34 J⋅sh = 6.626\times 10^{-34}\,\text{J}\cdot \text{s}.
    • Threshold frequency, f0=4.10×1014 Hzf_0 = 4.10 \times 10^{14}\,\text{Hz}.
    • Mass of electron, me=9.11×10−31 kgm_e = 9.11\times 10^{-31}\,\text{kg}.
    • Elementary charge, e=1.602×10−19 Ce = 1.602\times 10^{-19}\,\text{C} (useful for converting between joules and eV).

  2. Convert the given wavelength to frequency

    f  =  cλ  =  3.00×108 m/s585×10−9 m  ≈  5.13×1014 Hz.f \;=\; \frac{c}{\lambda} \;=\; \frac{3.00 \times 10^{8}\,\text{m/s}}{585\times 10^{-9}\,\text{m}} \;\approx\; 5.13 \times 10^{14}\,\text{Hz}.

  3. Compute the photon energy, EphotonE_\text{photon}

    Ephoton  =  h f  =  (6.626×10−34 J⋅s)×(5.13×1014 s−1).E_\text{photon} \;=\; h \, f \;=\; (6.626\times 10^{-34}\,\text{J}\cdot \text{s}) \times \bigl(5.13 \times 10^{14}\,\text{s}^{-1}\bigr).

    Numerically, this is roughly

    Ephoton  ≈  3.40×10−19 J.E_\text{photon} \;\approx\; 3.40 \times 10^{-19}\,\text{J}.

    (Alternatively, in electronvolts, using E≈1240 eV⋅nmλ(nm)E \approx \frac{1240\,\text{eV}\cdot\text{nm}}{\lambda(\text{nm})}, one gets Ephoton≈2.12 eV.E_\text{photon} \approx 2.12\,\text{eV}.)

  4. Compute the threshold energy, E0=h f0E_0 = h\,f_0

    E0  =  (6.626×10−34)×(4.10×1014)  ≈  2.72×10−19 J  ≈  1.70 eV.E_0 \;=\; (6.626\times 10^{-34}) \times \bigl(4.10 \times 10^{14}\bigr) \;\approx\; 2.72 \times 10^{-19}\,\text{J} \;\approx\; 1.70\,\text{eV}.

  5. Find the maximum kinetic energy of the ejected electrons
    By Einstein’s photoelectric equation,

    Kmax  =  Ephoton−E0.K_\text{max} \;=\; E_\text{photon} - E_0.

    In joules,

    Kmax  ≈  (3.40×10−19)  −  (2.72×10−19)  =  0.68×10−19 J  =  6.8×10−20 J.K_\text{max} \;\approx\; (3.40 \times 10^{-19}) \;-\; (2.72 \times 10^{-19}) \;=\; 0.68 \times 10^{-19}\,\text{J} \;=\; 6.8\times 10^{-20}\,\text{J}.

    (Or in eV, Kmax≈2.12−1.70=0.42 eV.K_\text{max} \approx 2.12 - 1.70 = 0.42\,\text{eV}.)

  6. Compute the maximum speed of the electrons
    Using Kmax=12mevmax⁡2K_\text{max} = \tfrac{1}{2} m_e v_{\max}^2:

    vmax⁡  =  2Kmaxme  =  2×6.8×10−209.11×10−31  ≈  1.2×105 m/s.v_{\max} \;=\; \sqrt{\frac{2K_\text{max}}{m_e}} \;=\; \sqrt{\frac{2 \times 6.8\times 10^{-20}}{9.11\times 10^{-31}}} \;\approx\; 1.2\times 10^{5}\,\text{m/s}.

Answers to the Questions

  1. What is the maximum speed of the ejected electrons?

    vmax⁡  ≈  1.2×105 m/s.\boxed{v_{\max} \;\approx\; 1.2 \times 10^{5}\,\text{m/s}.}

  2. What happens to the speed of electrons when the intensity of the yellow light increases? Why?

    • Short answer: The speed of the electrons does not increase when you only change the intensity of the light (keeping the same wavelength/frequency).
    • Reason: The speed (or kinetic energy) of ejected electrons depends on the photon energy, which is set by the frequency (or wavelength) of the light, not on its intensity.
      • Increasing intensity means more photons per second but does not change each photon’s energy.
      • Since the kinetic energy of each electron is determined by h f−h f0h\,f - h\,f_0, raising the intensity at the same frequency only increases the number of electrons ejected per second, not their maximum speed.

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