Question:
If the work functions of three photosensitive materials are 1 eV, 2 eV and 3 eV respectively, then the ratio of threshold frequencies of light that produce photoelectrons of maximum kinetic energy of 1 eV from each of them is
If the work functions of three photosensitive materials are 1 eV, 2 eV and 3 eV respectively, then the ratio of threshold frequencies of light that produce photoelectrons of maximum kinetic energy of 1 eV from each of them is
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$\nu \propto (\phi + K)$ — add both before comparing.
Updated On: May 2, 2026
- $1:2:3$
- $2:3:4$
- $1:1:1$
- $3:2:1$
- $4:3:2$
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The Correct Option is B
Solution and Explanation
Concept: Photoelectric equation
\[ h\nu = \phi + K_{\max} \] ---
Step 1: Given \[ K_{\max} = 1 \text{ eV (same for all)} \] ---
Step 2: For each material Material 1: \[ h\nu_1 = 1 + 1 = 2 \] Material 2: \[ h\nu_2 = 2 + 1 = 3 \] Material 3: \[ h\nu_3 = 3 + 1 = 4 \] ---
Step 3: Ratio of frequencies \[ \nu_1 : \nu_2 : \nu_3 = 2 : 3 : 4 \] --- Physical Insight:
• Higher work function → more energy needed
• Same kinetic energy → frequency must increase accordingly --- Final Answer: \[ \boxed{2:3:4} \]
\[ h\nu = \phi + K_{\max} \] ---
Step 1: Given \[ K_{\max} = 1 \text{ eV (same for all)} \] ---
Step 2: For each material Material 1: \[ h\nu_1 = 1 + 1 = 2 \] Material 2: \[ h\nu_2 = 2 + 1 = 3 \] Material 3: \[ h\nu_3 = 3 + 1 = 4 \] ---
Step 3: Ratio of frequencies \[ \nu_1 : \nu_2 : \nu_3 = 2 : 3 : 4 \] --- Physical Insight:
• Higher work function → more energy needed
• Same kinetic energy → frequency must increase accordingly --- Final Answer: \[ \boxed{2:3:4} \]
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