Question

A metal surface is illuminated by light of two different wavelengths \(207\text{ nm}\) and \(414\text{ nm}\) . The maximum speeds of photoelectrons corresponding to these wavelengths are \(u_1\) and \(u_2\) respectively with \(u_1 : u_2 = 2 : 1\). The work function of the metal is (\(hc = 1242\text{ eV nm}\))

• A. \(1.6\text{ eV}\)
• B. \(2.0\text{ eV}\)
• C. \(2.4\text{ eV}\)
• D. \(3.0\text{ eV}\)

Hint:

Shortcut: $K \propto v^2$

Answer 1

The Correct Option is B

Concept: Photoelectric equation: \[ K_{\max} = \frac{hc}{\lambda} - \phi \quad \text{and} \quad K \propto v^2 \]

Step 1: Given velocity ratio. \[ u_1 : u_2 = 2 : 1 \Rightarrow K_1 : K_2 = 4 : 1 \]

Step 2: Write equations. \[ K_1 = \frac{1242}{207} - \phi = 6 - \phi \] \[ K_2 = \frac{1242}{414} - \phi = 3 - \phi \]

Step 3: Use ratio. \[ \frac{6 - \phi}{3 - \phi} = 4 \] \[ 6 - \phi = 4(3 - \phi) \Rightarrow 6 - \phi = 12 - 4\phi \] \[ 3\phi = 6 \Rightarrow \phi = 2 \text{ eV} \]

Step 4: Conclusion.
Work function = $2.0$ eV Final Answer: Option (B)