Question

The photoelectric work function for a photosensitive material is 5.2 eV. The energy of the incident radiation for which the stopping potential is 6 V is

• A. 1.2 eV
• B. 5.6 eV
• C. 6 eV
• D. 10 eV
• E. 11.2 eV

Hint:

For photoelectric effect calculations, remember to convert the stopping potential to energy (in eV) by multiplying by the charge of an electron (approximately \(1.6 \times 10^{-19}\) Coulombs).

Answer 1

Answer: (E)

Step 1: Use the photoelectric equation: \( K_{{max}} = E - \phi \), where \(K_{{max}}\) is the maximum kinetic energy of the ejected electrons, \(E\) is the energy of the incident photons, and \(\phi\) is the work function. 
Step 2: The maximum kinetic energy can also be expressed as \( K_{{max}} = e \cdot V \), where \(V\) is the stopping potential and \(e\) is the elementary charge. 
Step 3: Setting \(e \cdot 6V = E - 5.2\) eV and solving for \(E\) gives \(E = 6 + 5.2 = 11.2\) eV. 
Step 4: Therefore, the energy of the incident radiation is 11.2 eV.