Question

The work function of photoelectric surface is 4.0 eV. Radiation of frequency \(3 \times 10^{15}\) Hz is incident on it. Calculate the maximum velocity of emitted photo-electron.

Hint:

When dealing with photoelectric effect problems, it's often convenient to work with energies in electron-volts (eV). Remember the conversion: 1 eV = \(1.6 \times 10^{-19}\) J. Only convert to SI units (Joules) when you need to calculate quantities like velocity or momentum that involve mass in kilograms.

Answer 1

Step 1: Understanding the Concept and Key Formula:
This problem involves the photoelectric effect. We need to use Einstein's photoelectric equation to find the maximum kinetic energy of the emitted electrons (photoelectrons) and then relate this kinetic energy to their maximum velocity.
Einstein's Photoelectric Equation: \[ K_{max} = hf - \phi \] where:
\( K_{max} \) is the maximum kinetic energy of the photoelectron.
\( h \) is Planck's constant (\(6.63 \times 10^{-34}\) J·s).
\( f \) is the frequency of the incident radiation.
\( \phi \) is the work function of the material.
The kinetic energy is related to velocity by \( K_{max} = \frac{1}{2} m_e v_{max}^2 \), where \(m_e\) is the mass of the electron (\(9.1 \times 10^{-31}\) kg).
Step 2: Calculating the Energy of the Incident Photon:
We are given:
Frequency, \( f = 3 \times 10^{15} \) Hz.
Work function, \( \phi = 4.0 \) eV.
First, calculate the energy of the incident photon (E = hf) in Joules: \[ E = (6.63 \times 10^{-34})(3 \times 10^{15}) \] \[ E = 19.89 \times 10^{-19} \, J \] Convert this to electron-volts (eV): \[ E = \frac{19.89 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 12.43 \, eV \] Step 3: Calculating the Maximum Kinetic Energy:
\[ K_{max} = E - \phi = 12.43 - 4.0 = 8.43 \, eV \] Step 4: Calculating the Maximum Velocity:
Convert kinetic energy back to Joules: \[ K_{max} = 8.43 \times (1.6 \times 10^{-19}) = 13.488 \times 10^{-19} \, J \] Now, solve for \( v_{max} \): \[ \tfrac{1}{2} m_e v_{max}^2 = 13.488 \times 10^{-19} \] \[ v_{max}^2 = \frac{2 \times 13.488 \times 10^{-19}}{9.1 \times 10^{-31}} \] \[ v_{max}^2 \approx 2.964 \times 10^{12} \] \[ v_{max} \approx 1.72 \times 10^6 \, m/s \] Step 5: Final Answer:
The maximum velocity of the emitted photoelectron is approximately \( 1.72 \times 10^6 \, m/s \).