Photoelectric Effect Speed Calculation

First, let us outline the key steps to solve this problem:

1. Identify the given data

   • Wavelength of incident light: $\lambda = 585\,\text{nm} = 585\times 10^{-9}\,\text{m}$
   • 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: $m_e = 9.11\times 10^{-31}\,\text{kg}$

2. Convert the incident light’s wavelength to frequency

   $$f \;=\; \frac{c}{\lambda} \;=\; \frac{3.00\times 10^8\,\text{m/s}}{585\times 10^{-9}\,\text{m}} \;\approx\; 5.13\times 10^{14}\,\text{Hz}.$$

3. Compute the photon energy of the incident light

   $$E_{\text{photon}} \;=\; h f \;=\; (6.626\times 10^{-34}\,\text{J}\cdot\text{s}) \times (5.13\times 10^{14}\,\text{s}^{-1}) \;\approx\; 3.40\times 10^{-19}\,\text{J}.$$

4. Compute the work function (threshold energy) from the threshold frequency

   $$W \;=\; h f_0 \;=\; (6.626\times 10^{-34}\,\text{J}\cdot\text{s}) \times (4.10\times 10^{14}\,\text{s}^{-1}) \;\approx\; 2.72\times 10^{-19}\,\text{J}.$$

5. Find the maximum kinetic energy
   The maximum kinetic energy $K_{\text{max}}$ of the ejected electrons is the photon energy minus the work function:

   $$K_{\text{max}} \;=\; E_{\text{photon}} \;-\; W \;=\; 3.40\times 10^{-19}\,\text{J} \;-\; 2.72\times 10^{-19}\,\text{J} \;=\; 0.68\times 10^{-19}\,\text{J}.$$

6. Compute the maximum speed of the ejected electrons
   Using the classical expression for kinetic energy,

   $$K_{\text{max}} \;=\; \tfrac{1}{2}\,m_e\,v_{\text{max}}^2,$$

   we solve for $v_{\text{max}}$:

   $$v_{\text{max}} \;=\; \sqrt{\frac{2\,K_{\text{max}}}{m_e}} \;=\; \sqrt{\frac{2 \times 0.68\times 10^{-19}\,\text{J}}{9.11\times 10^{-31}\,\text{kg}}}.$$

   Numerically,

   $$v_{\text{max}} \;\approx\; \sqrt{1.36\times 10^{-19} \;/\; 9.11\times 10^{-31}} \;\approx\; \sqrt{1.49\times 10^{11}} \;\approx\; 1.22\times 10^5\,\text{m/s}.$$

Thus,

$$\boxed{ v_{\text{max}} \;\approx\; 1.2\times 10^5\,\text{m/s}. }$$

---

What happens if the intensity of the yellow light is increased?

• Speed of the ejected electrons remains the same.
  The maximum kinetic energy (and thus the speed) of the photoelectrons depends only on the frequency (or wavelength) of the incident light, not on its intensity. Once the frequency is above the threshold, every photon carries the same energy. Hence, the speed of the ejected electrons is determined by the excess energy $\bigl[h(f - f_0)\bigr]$, which does not change if you merely turn up the brightness/intensity.

• Number of electrons ejected can increase.
  A higher intensity means more photons per second striking the photocathode. Consequently, more electrons can be ejected (assuming each photon has enough energy to exceed the work function), but each individual electron still has the same maximum speed determined by the light’s frequency.

Therefore, increasing the intensity of yellow light increases the number of photoelectrons ejected but does not increase their maximum speed.

Let’s zero in on the step involving the square root, since that’s where the discrepancy arises. The kinetic energy part and the division by the electron mass are correct, so let’s carefully re-compute:

---

1) Kinetic energy and the quantity under the square root

1. Kinetic energy

   $$K_{\text{max}} = 0.68 \times 10^{-19}\,\text{J}.$$

2. Double it (because $K = \tfrac12\,m\,v^2$)

   $$2\,K_{\text{max}} = 1.36 \times 10^{-19}\,\text{J}.$$

3. Divide by the electron mass

   $$\frac{2\,K_{\text{max}}}{m_e} = \frac{1.36 \times 10^{-19}\,\text{J}}{9.11 \times 10^{-31}\,\text{kg}} = \left(\frac{1.36}{9.11}\right) \times 10^{-19 - (-31)} = 0.149 \times 10^{12} = 1.49 \times 10^{11}.$$

Hence, the quantity inside the square root is $1.49 \times 10^{11}$.

---

2) Taking the square root correctly

$$v_{\text{max}} = \sqrt{1.49 \times 10^{11}} = \sqrt{1.49} \times \sqrt{10^{11}} = 1.22 \times 10^{\frac{11}{2}}.$$

But remember:

$$10^{\frac{11}{2}} \;=\; 10^{5.5} \;=\; 3.162 \times 10^{5}.$$

Therefore,

$$v_{\text{max}} = 1.22 \;\times\; 3.162 \times 10^{5} \;\approx\; 3.86 \times 10^{5}\,\text{m/s}.$$

So the corrected result is:

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

---

3) Why increasing intensity does not change this speed

• The maximum speed of photoelectrons depends on the photon energy above the work function. Once the frequency is fixed (i.e., we are still using 585 nm light), each photon has the same energy, so each ejected electron (assuming a single-photon interaction) has the same maximum kinetic energy regardless of how bright the light is.
• Increasing intensity simply means more photons per second; that can increase the number of electrons ejected per second, but the energy per electron (and hence its maximum speed) remains the same.