Question

Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Hint:

Shortcut:

Convert photon energy directly to eV
\( K_{\max} = h\nu - \phi \)
Stopping potential (in volts) = kinetic energy (in eV)

Answer 1

Concept: Photoelectric equation: \[ E = h\nu = \phi_0 + K_{\max} \] Stopping potential: \[ K_{\max} = eV_0 \] Constants: \[ h = 6.63 \times 10^{-34} \, \text{J·s}, \quad 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Step 1: Energy of photon. \[ E = h\nu = 6.63 \times 10^{-34} \times 6.4 \times 10^{14} \] \[ E = 4.24 \times 10^{-19} \, \text{J} \] Convert to eV: \[ E = \frac{4.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.65 \, \text{eV} \]
Step 2: Maximum kinetic energy. \[ K_{\max} = E - \phi_0 = 2.65 - 1.96 = 0.69 \, \text{eV} \]
Step 3: Stopping potential. \[ K_{\max} = eV_0 \] Since kinetic energy is in eV: \[ V_0 = 0.69 \, \text{V} \] Final Answers:

[(a)] Energy of photon = \( 2.65 \, \text{eV} \)
[(b)] Maximum kinetic energy = \( 0.69 \, \text{eV} \)
[(c)] Stopping potential = \( 0.69 \, \text{V} \)

Answer 2

We are given the work function of the metal, \( \phi_0 = 1.96\,\text{eV} \), and the frequency of the incident light, \( \nu = 6.4 \times 10^{14}\,\text{Hz} \). Instead of working directly in joules, this route first converts the frequency into a wavelength and then uses the handy eV-nanometre form of the photon energy formula.

Step 1: Convert frequency to wavelength.
\[ \lambda = \frac{c}{\nu} = \frac{3 \times 10^{8}}{6.4 \times 10^{14}} = 4.6875 \times 10^{-7}\,\text{m} = 468.75\,\text{nm} \]

Step 2: Find photon energy directly in eV.
Using the standard shortcut \( E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})} \), which comes from combining \( hc \) with the eV-joule conversion: \[ E = \frac{1240}{468.75} \approx 2.65\,\text{eV} \] This matches what a direct joule-based calculation would give, confirming the wavelength is computed correctly.

Step 3: Maximum kinetic energy of the photoelectrons.
By Einstein's photoelectric equation, the photon energy splits into the work needed to free the electron and the kinetic energy it carries away: \[ K_{\max} = E - \phi_0 = 2.65 - 1.96 = 0.69\,\text{eV} \]

Step 4: Stopping potential.
The stopping potential is the retarding voltage that just brings the fastest photoelectrons to rest, so \( eV_0 = K_{\max} \). Since \( K_{\max} \) is already expressed in eV, the numerical value of \( V_0 \) in volts equals the numerical value of \( K_{\max} \) in eV: \[ V_0 = 0.69\,\text{V} \]

So the photon energy is \( 2.65\,\text{eV} \), the maximum kinetic energy of the emitted electrons is \( 0.69\,\text{eV} \), and the stopping potential is \( 0.69\,\text{V} \).