Question

Photoemission of electrons occurs from a metal \((\phi_0=1.96\,\text{eV})\) when light of frequency \(6.4\times10^{14}\,\text{Hz}\) is incident on it. Calculate the maximum kinetic energy of the emitted electrons.

Hint:

For photoelectric effect problems, always use \[ K_{\max}=h\nu-\phi_0. \] If photon energy and work function are given in eV, perform the subtraction directly in eV.

Answer 1

Concept: According to Einstein's photoelectric equation, \[ E=h\nu=\phi_0+K_{\max}, \] where
• \(h\nu\) is the energy of the incident photon,
• \(\phi_0\) is the work function of the metal,
• \(K_{\max}\) is the maximum kinetic energy of the emitted photoelectrons. The maximum kinetic energy is therefore obtained by subtracting the work function from the photon energy.

Step 1: Write the given quantities. From part (a), \[ E=h\nu=2.65\,\text{eV}. \] Work function of the metal: \[ \phi_0=1.96\,\text{eV}. \]

Step 2: Apply Einstein's photoelectric equation. \[ K_{\max}=E-\phi_0. \] Substituting the values, \[ K_{\max} = 2.65-1.96. \] \[ K_{\max} = 0.69\,\text{eV}. \] Therefore, \[ \boxed{ K_{\max}=0.69\,\text{eV} }. \]

Step 3: Express the answer in SI units. Since \[ 1\,\text{eV}=1.6\times10^{-19}\,\text{J}, \] \[ K_{\max} = 0.69\times1.6\times10^{-19}. \] \[ K_{\max} = 1.104\times10^{-19}\,\text{J}. \] Hence, \[ \boxed{ K_{\max}=1.10\times10^{-19}\,\text{J} }. \] Final Answer: \[ \boxed{ K_{\max}=0.69\,\text{eV} } \] or \[ \boxed{ K_{\max}=1.10\times10^{-19}\,\text{J}. } \]