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Q11.2(c):
The work function of caesium metal is 2.14 eV. When light of frequency (6 \times 10^{14})Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum speed of the emitted photoelectrons?

Solution :

Step 1: Given Variables & Initial Setup

To determine the maximum speed ($v_{max}$) of the emitted photoelectrons, we utilize the Photoelectric Equation formulated by Albert Einstein. [Per the Law of Conservation of Energy in photoelectric emission].

  • Work function of Caesium ($\Phi_0$): $2.14\text{ eV}$
  • Frequency of incident light ($\nu$): $6 \times 10^{14}\text{ Hz}$
  • Planck’s constant ($h$): $6.626 \times 10^{-34}\text{ J}\cdot\text{s}$
  • Mass of an electron ($m_e$): $9.11 \times 10^{-31}\text{ kg}$
  • Conversion factor ($1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$)

Step 2: Conversion of Work Function to Joules

For consistency in SI units, we convert $\Phi_0$ from electron-volts to Joules:

$\Phi_0 = 2.14 \times 1.6 \times 10^{-19}\text{ J} = 3.424 \times 10^{-19}\text{ J}$

Step 3: Calculating Energy of Incident Photons ($E$)

Using the photon energy formula $E = h\nu$:

$E = (6.626 \times 10^{-34}\text{ J}\cdot\text{s}) \times (6 \times 10^{14}\text{ Hz})$

$E = 3.9756 \times 10^{-19}\text{ J}$

Step 4: Calculating Maximum Kinetic Energy ($K_{max}$ )

Applying the photoelectric equation: $K_{max} = E - \Phi_0$ [Per Einstein's Photoelectric Equation].

$K_{max} = 3.9756 \times 10^{-19}\text{ J} - 3.424 \times 10^{-19}\text{ J}$

$K_{max} = 0.5516 \times 10^{-19}\text{ J} = 5.516 \times 10^{-20}\text{ J}$

Step 5: Determining Maximum Speed ($v_{max}$)

Since $K_{max} = \frac{1}{2}m_e v_{max}^2$, we isolate $v_{max}$:

$v_{max} = \sqrt{\frac{2K_{max}}{m_e}}$

$v_{max} = \sqrt{\frac{2 \times 5.516 \times 10^{-20}}{9.11 \times 10^{-31}}}$

$v_{max} = \sqrt{1.211 \times 10^{11}}$

$v_{max} \approx 3.48 \times 10^{5}\text{ m/s}$

Visual Representation of Photoelectric Emission

Incident Photon ($h\nu$) Photoelectron ($v_{max}$) Caesium Surface ($\Phi_0 = 2.14 \text{ eV}$)

Final Solution: The maximum speed of the emitted photoelectrons is approximately $3.48 \times 10^5\text{ m/s}$.


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