First, let us outline the steps to solve for the maximum speed of the photoelectrons. Then we will discuss what happens when the light’s intensity changes.

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1. Given Information

• Wavelength of incident (yellow) light, $\lambda = 585\,\text{nm} = 5.85 \times 10^{-7}\,\text{m}.$
• Threshold frequency of the photocathode, f__{0} = 4.10 \times 10^{14}\,\text{Hz}.
• Speed of light, $c = 3.00 \times 10^{8}\,\text{m/s}.$
• Planck’s constant, $h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s}.$
• Electron mass, $m_e = 9.109 \times 10^{-31}\,\text{kg}.$

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2. Find the Frequency of the Incident Light

The frequency $f$ of light of wavelength $\lambda$ is

$$f = \frac{c}{\lambda}.$$

Substitute the values:

$$f = \frac{3.00 \times 10^{8}\,\text{m/s}}{5.85 \times 10^{-7}\,\text{m}} \approx 5.13 \times 10^{14}\,\text{Hz}.$$

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3. Calculate the Photon Energy Above the Threshold

1. Photon energy of the incident light: $$E_{\text{photon}} = h f = \bigl(6.626 \times 10^{-34}\,\text{J}\cdot\text{s}\bigr) \bigl(5.13 \times 10^{14}\,\text{Hz}\bigr) \approx 3.40 \times 10^{-19}\,\text{J}.$$
2. Work function ($\phi$) corresponds to the threshold frequency $f_0$: $$\phi = h f_0 = \bigl(6.626 \times 10^{-34}\bigr) \bigl(4.10 \times 10^{14}\bigr) \approx 2.71 \times 10^{-19}\,\text{J}.$$
3. Maximum kinetic energy of ejected electrons is the difference between the photon energy and the work function: $$K_{\max} = E_{\text{photon}} - \phi = 3.40 \times 10^{-19}\,\text{J} - 2.71 \times 10^{-19}\,\text{J} = 0.69 \times 10^{-19}\,\text{J}.$$ (Depending on rounding, you may see slightly different numbers; below we will show another more direct approach using the difference in frequency.)

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Alternative direct approach (using $f - f_0$ )

Often, we directly compute the energy difference via:

$$K_{\max} = h\,(f - f_0) = \bigl(6.626 \times 10^{-34}\bigr)\bigl(5.13 \times 10^{14} - 4.10 \times 10^{14}\bigr)\,\text{J} = h \times 1.03 \times 10^{14}\,\text{Hz}.$$ $$K_{\max} = 6.626 \times 10^{-34} \times 1.03 \times 10^{14} \approx 6.82 \times 10^{-20}\,\text{J}.$$

Depending on the precise rounding and intermediate steps, this yields a kinetic energy around $6.8 \times 10^{-20}\,\text{J}$, which is about $0.426\,\text{eV}.$

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4. Find the Maximum Speed of the Ejected Electrons

From classical kinetic energy,

$$K_{\max} = \tfrac{1}{2}\,m_e\,v_{\max}^2 \quad\Longrightarrow\quad v_{\max} = \sqrt{\frac{2\,K_{\max}}{m_e}}.$$

Using $K_{\max} \approx 6.82 \times 10^{-20}\,\text{J}$ and $m_e = 9.109 \times 10^{-31}\,\text{kg}$:

1. Compute the numerator: $$2\,K_{\max} \approx 2 \times 6.82 \times 10^{-20} = 1.36 \times 10^{-19}\,\text{J}.$$
2. Divide by the electron mass: $$\frac{1.36 \times 10^{-19}}{9.109 \times 10^{-31}} \approx 1.49 \times 10^{11}.$$
3. Take the square root: $$v_{\max} = \sqrt{1.49 \times 10^{11}} \approx 3.86 \times 10^{5}\,\text{m/s}.$$

Depending on rounding, you will often see $v_{\max} \approx 3.8\times 10^{5}\,\text{m/s}$.

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5. What Happens to the Electron Speed if the Light Intensity Increases?

• Key idea: The maximum kinetic energy (and thus speed) of the electrons depends on the frequency (or color) of the light, not on its intensity.
• If you increase the intensity but keep the same wavelength (585 nm), you are increasing the number of photons hitting the surface per second, but not their individual energy.
• Since the individual photon energy (determined by frequency $\nu$) stays the same, the excess energy above the work function stays the same, which means the speed of each ejected electron does not increase.
• What does increase with higher intensity at the same wavelength is the rate at which electrons are ejected (i.e., a greater photocurrent), but their maximum speed remains the same.

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Final Answers

1. Maximum speed of the ejected electrons:

   $$v_{\max} \approx 3.8\times 10^{5}\,\text{m/s}.$$

2. Effect of increasing intensity:

   • The maximum speed of the electrons does not change because it depends on the frequency (color) of the light, not on the intensity.
   • Increasing the intensity at the same wavelength merely increases the number of electrons ejected, but each electron’s kinetic energy (and hence speed) stays the same.

Let’s do the final step with careful attention to the arithmetic. The potential concern is in taking the square root of the quantity $\frac{2K_{\max}}{m_e}$.

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Recap of the Key Quantities

1. Planck’s constant: $$h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s}$$
2. Speed of light: $$c = 3.00 \times 10^{8}\,\text{m/s}$$
3. Electron mass: $$m_e = 9.109 \times 10^{-31}\,\text{kg}$$
4. Threshold frequency: $$f_0 = 4.10 \times 10^{14}\,\text{Hz}$$
5. Incident wavelength: $$\lambda = 585\,\text{nm} = 5.85 \times 10^{-7}\,\text{m}$$ so the incident frequency is $$f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{5.85 \times 10^{-7}} \approx 5.13 \times 10^{14}\,\text{Hz}.$$

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1. Compute the Photon Energy and the Work Function

• Photon energy of incident light:

  $$E_{\text{photon}} = h f = (6.626 \times 10^{-34}\,\text{J}\cdot\text{s}) (5.13 \times 10^{14}\,\text{Hz}) \approx 3.40 \times 10^{-19}\,\text{J}.$$

• Work function ($\phi$) of the metal, given by the threshold frequency $f_0$:

  $$\phi = h f_0 = (6.626 \times 10^{-34}\,\text{J}\cdot\text{s}) (4.10 \times 10^{14}\,\text{Hz}) \approx 2.71 \times 10^{-19}\,\text{J}.$$

• Maximum kinetic energy of ejected electrons:

  $$K_{\max} = E_{\text{photon}} - \phi \approx (3.40 \times 10^{-19}) - (2.71 \times 10^{-19}) = 0.69 \times 10^{-19}\,\text{J} = 6.9 \times 10^{-20}\,\text{J}.$$

  (Or more precisely, using the direct frequency difference $f - f_0 = 1.03 \times 10^{14}\,\text{Hz}$: $\,K_{\max} = h(f - f_0) \approx 6.82\times10^{-20}\,\text{J}.$)

Let’s take $K_{\max} \approx 6.82 \times 10^{-20}\,\text{J}$ for a bit more precise arithmetic.

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2. Compute $v_{\max}$ Carefully

The classical relation for the electron’s maximum kinetic energy is

$$K_{\max} = \tfrac{1}{2} m_e v_{\max}^2 \quad\Longrightarrow\quad v_{\max} = \sqrt{\frac{2K_{\max}}{m_e}}.$$

Step-by-step:

1. Double the kinetic energy:

   $$2\,K_{\max} = 2 \times 6.82 \times 10^{-20}\,\text{J} = 1.364 \times 10^{-19}\,\text{J}.$$

2. Divide by the electron mass:

   $$\frac{1.364 \times 10^{-19}}{9.109 \times 10^{-31}} = 1.364 \,\big/\, 9.109 \times 10^{-19 + 31} = 0.1497 \times 10^{12} = 1.497 \times 10^{11}.$$

3. Take the square root:

   $$\sqrt{1.497 \times 10^{11}} = \sqrt{1.497}\,\times\,\sqrt{10^{11}} \approx (1.22)\,\times\,(10^{5.5}).$$
   • $\sqrt{1.497} \approx 1.22$ (since $\sqrt{1.44} = 1.20$ and $\sqrt{1.49} \approx 1.22$).
   • $\sqrt{10^{11}} = 10^{11/2} = 10^{5.5} \approx 3.16 \times 10^5.$

4. Multiply:

   $$v_{\max} \approx 1.22 \;\times\; 3.16\times 10^5 \approx 3.86 \times 10^5\,\text{m/s}.$$

Depending on your rounding at each step, you may see $3.86\times10^5$, $3.87\times10^5$, or $3.9\times10^5$ m/s, but they are all essentially the same to 2–3 significant figures.

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3. What If You Increase the Intensity of the Same Yellow Light?

• Key fact: The kinetic energy (and thus maximum speed) of the emitted electrons depends on the frequency of the light, not on its intensity.
• If you increase intensity (i.e., make the light brighter) but keep the same wavelength (585 nm), the energy per photon remains the same.
• Because each electron’s maximum kinetic energy is still $h(f - f_0)$, the electron speed does not increase when intensity goes up.
• Instead, you will emit more electrons per second (higher photocurrent), but each electron’s speed remains the same as long as the frequency remains at 5.13 × 10^14 Hz.

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Final Answer

1. Maximum speed of the ejected electrons is about

$$\boxed{v_{\max} \approx 3.9\times 10^5\text{ m/s}.}$$

2. If the intensity of the yellow light increases (but the color stays the same), the maximum speed of the electrons does not change. Their speed depends only on the photon frequency (i.e., the color), not on how many photons arrive per second. Increasing intensity increases the number of electrons ejected per second, but not their individual kinetic energies.