Question

Stopping potential is \(2.5\,V\) for light of wavelength \(400\,nm\). Find the work function of the metal.

• A. \(0.6\,eV\)
• B. \(2.5\,eV\)
• C. \(3.1\,eV\)
• D. \(3.7\,eV\)

Hint:

For wavelength given in nanometers, \[ E(eV)=\frac{1240}{\lambda(nm)} \] This shortcut is extremely useful in photoelectric effect problems.

Answer 1

The Correct Option is A

Concept: According to Einstein's photoelectric equation, \[ h\nu=\phi+K_{\max} \] where
• \(h\nu\) = energy of incident photon
• \(\phi\) = work function of metal
• \(K_{\max}\) = maximum kinetic energy of emitted electrons Also, \[ K_{\max}=eV_s \] where \(V_s\) is the stopping potential.

Step 1: Calculate photon energy. Using \[ E=\frac{1240}{\lambda(nm)} \] \[ E=\frac{1240}{400} \] \[ E=3.1\,eV \]

Step 2: Determine maximum kinetic energy. Given stopping potential \[ V_s=2.5V \] Hence \[ K_{\max}=2.5\,eV \]

Step 3: Apply Einstein's equation. \[ \phi=E-K_{\max} \] \[ \phi=3.1-2.5 \] \[ \phi=0.6\,eV \] Final Answer: \[ \boxed{\phi=0.6\,eV} \] Hence the correct option is \[ \boxed{(A)} \]